a) Show that the equations \begin{align*} x^3-y^2+2u-v&=1,\\ x^2+y^3+u-v&=2 \end{align*} can be solved simultaneously for functions $u=g(x,y),\ v=h(x,y)$ in a neighbourhood of $(1,1)$ such that $g(1,1) = h(1,1) = 1$.
b) Calculate $\frac{dg}{dx}$ and $\frac{dh}{dy}$ at the point $(1,1)$.
For part (a), I got the answer $-1$. For part (b), my $\frac{dg}{dx}$ is $2x - 3x^2$ and $\frac{dh}{dy}$ is $6y^2 + 2y$.
Please let me know if I am wrong and why; and if i'm correct, please let me know too! Much appreciated!
For part (a) you should be finding the functions $g(x,y)$ and $h(x,y)$ by solving for $u$ and $v$ (do this either by the substitution or elimination methods you would have learnt). The answers should involve $x$ and $y$.
For part (b) you then differentiate the above functions with respect to the appropriate variables. Your current answer is close, but not quite right.