Given the surface $z = f(x,y)$, with parameterization $x = u + v^2$, $y = u^2 - v^3$, $z = 2uv$ near the point $(3,3,4)$ which corresponds to the point $(2,1)$ in the $uv$-plane, find $\frac{\partial^2 f}{\partial x \partial y}(3,3)$.
I'm familiar with using Jacobians to find first partial derivatives, but not how to find higher order partial derivatives of multivariate implicit functions.
By the Inverse Function Theorem, the equations $x=u+v^2$ and $y=u^2-v^3$ locally define $u$ and $v$ as $C^1$ functions of $x$ and $y$ near the point in question, since $x(u,v)=u+v^2$ and $y(u,v)=u^2-v^3$ are functions of class $C^1$ and the Jacobian determinant $$ \begin{vmatrix} x_u & x_v \\ y_u & y_v \end{vmatrix} = \begin{vmatrix} 1 & 2v \\ 2u & -3v^2 \end{vmatrix} = - (3v^2 + 4uv) $$ is nonzero at the point $(u,v)=(2,1)$.
The partial derivatives of the inverse functions $u(x,y)$ and $v(x,y)$ are given by $$ \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix} = \begin{pmatrix} x_u & x_v \\ y_u & y_v \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 2v \\ 2u & -3v^2 \end{pmatrix}^{-1} = \frac{1}{3v^2 + 4uv} \begin{pmatrix} 3v^2 & 2v \\ 2u & -1 \end{pmatrix} , $$ from which we read off $$ u_x = \frac{3v^2}{3v^2 + 4uv} , $$ and so on.
The formulas $z=f(x,y)$ and $z=2uv$ say that $f$ is simply $$ f(x,y) = 2 \,u(x,y) \, v(x,y) , $$ so $$ f_{xy} = 2 (u_{xy} v + u_x v_y + u_y v_x + u v_{xy}) . $$ In this expression, we already know everything except $u_{xy}$ and $v_{xy}$. From the chain rule, we get $$ u_{xy} = (u_x)_{y} = (u_x)_u \, u_y + (u_x)_v \, v_y = \frac{\partial}{\partial u} \left( \frac{3v^2}{3v^2 + 4uv} \right) \, u_y + \frac{\partial}{\partial v} \left( \frac{3v^2}{3v^2 + 4uv} \right) \, v_y = \cdots $$ and similarly for $v_{xy}$. I'll leave it to you to complete the computation, and to insert $(u,v)=(2,1)$ to get $f_{xy}(3,3)$.
