I need to find the tangent plane to the function $h(x,y)=f(x^2y-2x,x^2-y^4-1)$ at $(1,1,h(1,1))$, where $f$ is a two variable differentiable function such that, $\nabla f(1,1)=(-\sqrt{2},-6)$; $\nabla f(-1,-1)=(-2,3)$; $f(1,1)=2$ and $f(-1,-1)=1$.
My attempt:
I defined $u(x,y)=x^2y-2x$ and $v(x,y)=x^2-y^4-1$. Notice that, $$h(1,1)=f(u(1,1),v(1,1))=f(-1,-1)=1.$$ Also by using the chain rule we have, $$\frac{\partial h}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x},$$ and, $$\frac{\partial h}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}.$$ How can I evaluate $\frac{\partial h}{\partial x}(1,1)$ and $\frac{\partial h}{\partial y}(1,1)$ by using the expressions above and the given values for $f$ and $\nabla f$?
hint: $\displaystyle \frac{\partial h}{\partial x} = \nabla f \cdot \left\langle \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x} \right\rangle$, hence $\displaystyle \frac{\partial h}{\partial x}(1,1) = \nabla f (1,1) \cdot \left\langle \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x} \right\rangle\biggr\lvert_{(1,1)}$
Similarly: $\displaystyle \frac{\partial h}{\partial y} = \nabla f \cdot \left\langle \frac{\partial u}{\partial y}, \frac{\partial v}{\partial y} \right\rangle$
Also $h(1,1)= f(-1,-1)$
Of course, once you evaluate everything at the corresponding point you'll have the given values.