Consider the equation $e^{xz}+y-z=e$. Using the implicit function theorem shows $z$ is a smooth function of $x,y$ about $(1,1,1)$. I needed to calculate a directional derivative of $z$ at $(1,1)$ and managed that using the implicit function theorem to recover the gradient of $z$.

Now I'm asked whether the partial derivatives of $z$ are symmetric about $(1,1)$ and furthermore, I need to calculate them.

I think the partial derivatives *are* symmetric because the original function $e^{xz}+y-z=e$ is smooth, which means so is $z=z(x,y)$. I don't understand however how to find second order derivatives. The "formula" $$\frac{\partial z}{\partial x}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}$$ (read with matrix inverse instead of quotient in matrix case) does not really make sense before it's evaluated at a point, since the RHS has additional variables.

So how to find $\frac{\partial ^2z}{\partial x\partial y}(1,1)$?