Trouble with partial derivatives I've no clue how to get started .I'am unable to even understand what the hint is saying.I need your help please.
Given $$u = f(ax^2 + 2hxy + by^2), \qquad v = \phi (ax^2 + 2hxy + by^2),$$ then prove that
$$\frac{\partial }{\partial y} \left ( u\frac{\partial u }{\partial x} \right ) = \frac{\partial }{\partial x}\left ( u \frac{\partial v}{\partial y} \right ).$$
Hint. Given $$u = f(z),v = \phi(z), \text{where} z = ax^2 + 2hxy + by^2$$
 A: I think that the hint and chain rule are more than enough. There may be a typo in your question since in the RHS you have both $u,v$ in the derivation and in the LHS you have only $u$.
$$u=f(z) \Rightarrow \frac{\partial u}{\partial x}=f'(z)\cdot \frac{\partial z}{\partial x}=f'(z)\cdot(2ax +2hy)$$
Now to compute everything in the LHS you just plug in $v$ and the previous result and use the product rule.
$$ \frac{\partial}{\partial y}\left(f(z) \cdot f'(z) \cdot (2ax+2hy)\right),$$
and again, use chain rule whenever necessary.
A: I recommend you to use the chain rule, i.e. given functions $f:U\mathbb{R}^n\rightarrow V\subset\mathbb{R}^m$ differenciable on $x$ and $g:V_0\subset V\subset\mathbb{R}^m\rightarrow W\subset\mathbb{R}^p$ differenciable on $f(x)$ we have
$$D_x(g\circ f)=D_{f(x)}(g)D_x(f)$$
where $D_x(f)$ represents the Jacobian matriz of $f$ at $x$. 
In your particular case when $n=2$ and $m=p=1$, we have for each coordinate that
$$\left.\frac{\partial g\circ f}{\partial x_i}\right|_{(x_1,x_2)}=\left.\frac{d\,g}{dx}\right|_{f(x_1,x_2)}\left.\frac{\partial f}{\partial x_i}\right|_{(x_1,x_2)}$$
