Q:

Calculus Partial derivative computation

I have the following systems of equation
$$x^5v^2+2y^3u = 3 \\3yu - xuv^3 = 2$$  
I need to show that this system defines v and u implicitly as functions of x and y near the point $(x,y,u,v) = (1,1,1,1)$. Compute all their partial derivatives at $(x,y)=(1,1)$ for local functions $v$ and $u$.
Not sure how to start this so any help is appreciated. 

A:

Consider $\vec{x}=\binom{u}{v}$ and $\vec{y} = \binom{x}{y}$. Now let $F(\vec{x}, \vec{y}) = \binom{x^5v^2+2y^3u-3}{3yu-xuv^3-2}=0$. Take the derivative w.r.t $\vec{y}$ to get 
$$F_{\vec{y}}(\vec{x},\vec{y})= \begin{pmatrix}5x^4v^2 & 6y^2u\\-uv^3 &3u\end{pmatrix}$$
Now evaluate in $\vec{x_0}= \binom{1}{1},\vec{y_0} = \binom{1}{1}$. Then 
$$F_{\vec{y}}(\vec{x_0},\vec{y_0})= \begin{pmatrix}5 & 6\\-1 &3\end{pmatrix}$$ 
that has determinant $det F_{\vec{y}}(\vec{x_0},\vec{y_0}) = 21 \neq 0$. What does the Implicit Function Theorem tell you? 