Implicit function theorem exercise problem By eliminating $u$ from the equations, $x=u+v$ and $y=uv^2$, we get that $F(x,y,v)=0$. Where $v$ implicitly defines $x$ as function of $y$ ($v=h(x,y)$). Prove that $\dfrac{\partial h}{\partial x}=\dfrac{h(x,y)}{3h(x,y)-2x}$.
Honestly I have no idea how to start. First, how do we get $F(x,y,v)=0$? What does eliminating means? To give a specific value to $v$? (I wrote it out as the problem was given)
 A: The statement of the exercise seems a bit strange to me. Normally there is a given point $(x_0, y_0, v_0) \in \mathbb{R}^3$ where you have to check the condition of the implicit function theorem in order to conclude the existence of your function $h(x,y)$ on an appropriate neighborhood around that point.
However, we can start the calculation anyways and say later at which points the result holds, i.e. where the theorem is applicable. So, you can write $u = x - v$ and plug it into $y = uv^2$, which gives you $y = xv^2 - v^3$ or equivalently $xv^2 - v^3 - y = 0$ . Then we can define 
\begin{equation}
F: \mathbb{R}^2 \times\mathbb{R} \to \mathbb{R}, \, F(x, y, v) = xv^2 - v^3 - y
\end{equation}
Now, for applying the implicit function theorem we need $\frac{\partial F}{\partial v} = 2xv -3v^2 \neq 0$, which is the case if $v \neq 0$ or $3v \neq 2x$. Assuming this, we conclude the existence of a function $h(x, y)$ on an appropriate neighborhood such that
\begin{equation}
F(x, y, v) = 0 \Leftrightarrow v = h(x,y)
\end{equation}
for all $(x, y, v)$ in that neighborhood.
Next, the theorem gives us a formula for calculating the derivative of $h$. Writing $w := (x,y)$, we have
\begin{equation}
\frac{\partial h}{\partial w}(w) = -(\frac{\partial F}{\partial v}(w,h(w)))^{-1} \cdot \frac{\partial F}{\partial w}(w,h(w)),
\end{equation}
where by $\frac{\partial F}{\partial v}$ the normal partial derivative of $F$ with respect to $v$, and by $\frac{\partial h}{\partial w}$ resp. $\frac{\partial F}{\partial w}$ the matrix $(\frac{\partial h}{\partial x}, \, \frac{\partial h}{\partial y})$ resp. $(\frac{\partial F}{\partial x}, \, \frac{\partial F}{\partial y})$ is meant. If we consider $F(w, h(w)) = xh(w)^2 - h(w)^3 - y$, we get


*

*$-(\frac{\partial F}{\partial v}(w, h(w))^{-1} = -(2xh(w) - 3h(w)^2)^{-1}$

*$\frac{\partial F}{\partial w}(w, h(w)) = (h(w)^2, \, -1)$


Since we are only interested in the partial derivative of $h$ with respect to $x$, we only have to consider the first component of $\frac{\partial F}{\partial w}$. So, together with the general formula for the derivative above we get
\begin{equation}
\frac{\partial h}{\partial x} = -\frac{h(w)^2}{(2xh(w) - 3h(w)^2)} = \frac{h(w)}{3h(w) - 2x}
\end{equation}
as desired. 
For a more exact statement of the implicit function theorem and on how to use it in general, you can also have a look at my answer here. 
A: In order to eliminate $u$ plug$u=x-v$ from the first equation into the second equation, and obtain $y=(x-v)v^2$, or
$$F(x,y,v):=v^3-v^2x +y=0\ .\tag{1}$$
This is a "law" connecting the three a priori independent variables $x$, $y$, $v$. Assume that the point $(x_0,y_0,v_0)$ fulfills this law, i.e., $F(x_0,y_0,_0)=0$. Then under suitable technical assumptions there is a box-window with center $(x_0,y_0,v_0)$ such that within this window all points satisfying $(1)$ are given by a $C^1$-function $(x,y)\mapsto v=h(x,y)$, whereby $h(x_0,y_0)=v_0$. We therefore have
$$h^3(x,y)-h^2(x,y)\,x+y\equiv0\tag{2}$$
in a neighborhood of $(x_0,y_0)$. Differentiating $(2)$ partially  with respect to $x$ gives
$$\bigl(3h^2(x,y)-2h(x,y) x\bigr)h_x-h^2(x,y)\equiv0\ ,$$
and solving for $h_x$ we obtain
$${\partial h\over\partial x}(x,y)={h(x,y)\over 3h(x,y)-2x}\ .$$
The "technical assumptions" here are that ${\partial F\over\partial v}(x_0,y_0,v_0)=v_0(3v_0-2x_0)\ne0$.
