Implicit Function Theorem to solve $\frac{\partial u}{\partial x}$ when $f\big(x,y,u(x,y),v(x,y)\big)$ mapped to $\mathbb{R^2}$ I am given $2$ equations 
$$f_1=x^2-y^2-u^3+v^2+4=0$$
$$f_2=2xy+y^2-2u^2+3v^4+8=0$$
and asked to show that we can solve the equations for $u$ and $v$ in terms of $x$ and $y$ in a neighbhourood of the solution $(2,-1,2,1)$.  I have applied the implicit function theorem and showed that $det\bigg(\frac{\partial f_1,f_2}{\partial_u,v}\bigg)$ at $(2,-1,2,1)$ is non-zero.
My Question
They then ask to solve for $\frac{\partial u}{\partial x}$.  This is what I'm having trouble with.  I know that when we, for e.g, have $f(x,g(x))$ mapped to one function then we can solve for $\frac{dy}{x}$ by computing $\frac{dy}{x}=-\frac{f_x}{f_y}$.  I tried applying the same definition to this problem but had some difficulty.  I know that I can derive $f_1$ with respect to $x$ thus getting $\frac{\partial f_1}{\partial x}=2x-3u^2\frac{\partial u}{\partial x} +2v\frac{\partial v}{\partial x}$ and can do the same for $f_2$.  And from there I can eliminate $\frac{\partial v}{\partial x}$ and isolate for $\frac{\partial u}{\partial x}$.  Is this the only procedure to solve for $\frac{\partial u}{\partial x}$ or is there a formula like in the case for one component function.
Thanks
 A: Such a formula exists and, as in the scalar case, it follows from the chain rule. Let $x\in\mathbb R^n$, $y\in\mathbb R^m$, $F(x,y):\;\mathbb R^n\times \mathbb R^m\to \mathbb R^m$. Suppose that in some neiborghood of the point $(x,y):\;F(x,y)=0$ $F$ is continuously differentiable and a continuously differentiable implicit function $y=\phi(x)$ exists.
Differentiating the equality $F(x,\phi(x))=0$ with respect to the variables $x$, one can obtain
$$
\frac{\partial F}{\partial x}\frac{\partial x}{\partial x}+
\frac{\partial F}{\partial y}\frac{\partial \phi(x)}{\partial x}=0
$$
$$
\frac{\partial F}{\partial x}+
\frac{\partial F}{\partial y}\frac{\partial \phi(x)}{\partial x}=0
$$
and, finally, if $\det \left( \frac{\partial F}{\partial y} \right)\ne 0$,
$$
\frac{\partial \phi(x)}{\partial x}=-
\left( \frac{\partial F}{\partial y} \right)^{-1}
\frac{\partial F}{\partial x}.
$$
For example, for the equations in your question one obtains (maybe, using some symbolic math software)
$$
\frac{\partial F}{\partial (x,y)}=
\left(\begin{array}{cc} 2 x & - 2 y\\ 2y & 2 x + 2 y \end{array}\right),
\quad
\frac{\partial F}{\partial (u,v)}=
\left(\begin{array}{cc} - 3 u^2 & 2 v\\ - 4 u & 12 v^3 \end{array}\right),
$$
$$
\left(\frac{\partial F}{\partial (u,v)}\right)^{-1}=
\left(\begin{array}{cc} \frac{3v^2}{2u - 9 u^2 v^2} & -\frac{1}{2 \left(2 u - 9 u^2 v^2\right)}\\ \frac{1}{2 v - 9 u v^3} & -\frac{3 u}{4 \left(2 v - 9 u v^3\right)} \end{array}\right),
$$
$$
\left(\begin{array}{cc}
\frac{\partial u}{\partial x} &  \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x} &  \frac{\partial v}{\partial y}\\
 \end{array}\right)=
-\left(\begin{array}{cc} \frac{3v^2}{2u - 9 u^2 v^2} & -\frac{1}{2 \left(2 u - 9 u^2 v^2\right)}\\ \frac{1}{2 v - 9 u v^3} & -\frac{3 u}{4 \left(2 v - 9 u v^3\right)} \end{array}\right)
\left(\begin{array}{cc} 2 x & - 2 y\\ 2y & 2 x + 2 y \end{array}\right)
$$
$$
=\left(\begin{array}{cc} -\frac{y - 6 v^2 x}{u\, \left(9 u v^2 - 2\right)} & -\frac{6 y v^2 + x + y}{u \left(9 u v^2 - 2\right)}\\ \frac{4 x - 3 u y}{2 v \left(9 u v^2 - 2\right)} & -\frac{4 y + 3 u x + 3 u y}{2 v \left(9 u v^2 - 2\right)} \end{array}\right)
$$
