I'm not sure I follow why your solution is \( C^\infty \)
The function is continuous, of course:
\( \begin{eqnarray}
\lim_{y \to {\small 1^{-}}} F(x,y, 0) & = & x^y \\
& = & x^1 \\
\\[10pt]
\\
& = & x \\
\\[15pt]
\\
\lim_{y \to {\small 1^{+}}} F(x,y-1, 1) & = & x^{x^{y-1}} \\
& = & x^{x^0} \\
\\[5pt]
\\
& = & x^1 \\
\\[10pt]
\\
& = & x
\end{eqnarray}
\)
And for base e, the first derivative is continuous as well:
\( \begin{eqnarray}
\lim_{y \to {\small 1^{-}}} D_y F(x,y, 0) & = & D_y x^y \\
& = & x^y \ln(x) \\
& = & x^1 \ln(x) \\
& = & x \ln(x) \\
\\[10pt]
\\
\lim_{y \to {\small 1^{+}}} D_y F(x,y-1, 1) & = & D_y \left(x^{x^{y-1}}\right) \\
& = & \left(x^{x^{y-1}}\right) \ln(x) D_x (x^{y-1}) \\
& = & \left(x^{x^{y-1}}\right) \left(x^{y-1}\right) \left(\ln(x)\right)^2 \\
& = & \left(x^{x^0}\right) (x^0) \left(\ln(x)\right)^2 \\
& = & x \left(\ln(x)\right)^2
\end{eqnarray} \)
However, even at the second derivative, we begin to see problems:
\( \begin{eqnarray}
\lim_{y \to {\small 1^{-}}} D_y^2 F(x,y, 0) & = & D_y^2 x^y \\
& = & D_y \left( x^y\right) \ln(x) \\
& = & x^y \left(\ln(x)\right)^2 \\
& = & x^1 \left(\ln(x)\right)^2 \\
& = & x \left(\ln(x)\right)^2 \\
\\[10pt]
\\
\lim_{y \to {\small 1^{+}}} D_y^2 F(x,y-1, 1) & = & D_y^2 \left(x^{x^{y-1}}\right) \\
& = & D_y \left( \left(x^{x^{y-1}}\right)\left( x^{y-1}\right)\right) \left(\ln(x)\right)^2 \\
& = & \left(\left( \left[D_y \left(x^{x^{y-1}}\right)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left[D_y \left( x^{y-1}\right)\right]\right)\right) \left(\ln(x)\right)^2 \\
& = & \left(\left( \left[x^{x^{y-1}} \ln(x) D_y \left(x^{y-1}\right)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left( x^{y-1}\right) \ln(x) \right)\right) \left(\ln(x)\right)^2 \\
& = & \left(\left( \left[x^{x^{y-1}} \ln(x) \left(x^{y-1}\right)\ln(x)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left( x^{y-1}\right) \ln(x) \right)\right) \left(\ln(x)\right)^2 \\
& = & \left( \left(x^{x^{y-1}}\right) \left(x^{y-1}\right) \left(\ln(x)\right)^3 \right) \left(\left(x^{y-1}\right)\ln(x) + 1\right) \\
& = & \left( \left(x^{x^0}\right) \left(x^0\right) \left(\ln(x)\right)^3 \right) \left(\left(x^0\right)\ln(x) + 1\right) \\
& = & \left( \left(x^1\right) \left(1\right) \left(\ln(x)\right)^3 \right) \left(\left(1\right)\ln(x) + 1\right) \\
& = & \left( x \left(\ln(x)\right)^3 \right) \left(\ln(x) + 1\right) \\
\end{eqnarray} \)
For base e, we get a second derivative of 1 from the left, but 2 from the right. Due to the product rule, the second and higher derivatives will get quite complex, preventing a simple solution.
The function is continuous, of course:
\( \begin{eqnarray}
\lim_{y \to {\small 1^{-}}} F(x,y, 0) & = & x^y \\
& = & x^1 \\
\\[10pt]
\\
& = & x \\
\\[15pt]
\\
\lim_{y \to {\small 1^{+}}} F(x,y-1, 1) & = & x^{x^{y-1}} \\
& = & x^{x^0} \\
\\[5pt]
\\
& = & x^1 \\
\\[10pt]
\\
& = & x
\end{eqnarray}
\)
And for base e, the first derivative is continuous as well:
\( \begin{eqnarray}
\lim_{y \to {\small 1^{-}}} D_y F(x,y, 0) & = & D_y x^y \\
& = & x^y \ln(x) \\
& = & x^1 \ln(x) \\
& = & x \ln(x) \\
\\[10pt]
\\
\lim_{y \to {\small 1^{+}}} D_y F(x,y-1, 1) & = & D_y \left(x^{x^{y-1}}\right) \\
& = & \left(x^{x^{y-1}}\right) \ln(x) D_x (x^{y-1}) \\
& = & \left(x^{x^{y-1}}\right) \left(x^{y-1}\right) \left(\ln(x)\right)^2 \\
& = & \left(x^{x^0}\right) (x^0) \left(\ln(x)\right)^2 \\
& = & x \left(\ln(x)\right)^2
\end{eqnarray} \)
However, even at the second derivative, we begin to see problems:
\( \begin{eqnarray}
\lim_{y \to {\small 1^{-}}} D_y^2 F(x,y, 0) & = & D_y^2 x^y \\
& = & D_y \left( x^y\right) \ln(x) \\
& = & x^y \left(\ln(x)\right)^2 \\
& = & x^1 \left(\ln(x)\right)^2 \\
& = & x \left(\ln(x)\right)^2 \\
\\[10pt]
\\
\lim_{y \to {\small 1^{+}}} D_y^2 F(x,y-1, 1) & = & D_y^2 \left(x^{x^{y-1}}\right) \\
& = & D_y \left( \left(x^{x^{y-1}}\right)\left( x^{y-1}\right)\right) \left(\ln(x)\right)^2 \\
& = & \left(\left( \left[D_y \left(x^{x^{y-1}}\right)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left[D_y \left( x^{y-1}\right)\right]\right)\right) \left(\ln(x)\right)^2 \\
& = & \left(\left( \left[x^{x^{y-1}} \ln(x) D_y \left(x^{y-1}\right)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left( x^{y-1}\right) \ln(x) \right)\right) \left(\ln(x)\right)^2 \\
& = & \left(\left( \left[x^{x^{y-1}} \ln(x) \left(x^{y-1}\right)\ln(x)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left( x^{y-1}\right) \ln(x) \right)\right) \left(\ln(x)\right)^2 \\
& = & \left( \left(x^{x^{y-1}}\right) \left(x^{y-1}\right) \left(\ln(x)\right)^3 \right) \left(\left(x^{y-1}\right)\ln(x) + 1\right) \\
& = & \left( \left(x^{x^0}\right) \left(x^0\right) \left(\ln(x)\right)^3 \right) \left(\left(x^0\right)\ln(x) + 1\right) \\
& = & \left( \left(x^1\right) \left(1\right) \left(\ln(x)\right)^3 \right) \left(\left(1\right)\ln(x) + 1\right) \\
& = & \left( x \left(\ln(x)\right)^3 \right) \left(\ln(x) + 1\right) \\
\end{eqnarray} \)
For base e, we get a second derivative of 1 from the left, but 2 from the right. Due to the product rule, the second and higher derivatives will get quite complex, preventing a simple solution.