Partial derivatives. Suppose 
$$f(x+y, x^2 +xy + z^2) = 0.$$ 
Show that
$$x + y = 2z\left(\frac{\partial z}{\partial y}-\frac{\partial z}{\partial x}\right).$$
Please help I don't know where to start!

Let's assume that 
$$k(x,y)=x+y$$
$$l(x,y,z(x,y))=x^2+xy+z^2$$
Then the function becomes
$$f\bigg(k(x,y),l\big(x,y,z(x,y)\big)\bigg)=0$$
If you take the total differential
$$df=\bigg(\frac{\partial f}{\partial k}\frac{\partial k}{\partial x}+\frac{\partial f}{\partial l}\frac{\partial l}{\partial x}\bigg)dx+\bigg(\frac{\partial f}{\partial k}\frac{\partial k}{\partial y}+\frac{\partial f}{\partial l}\frac{\partial l}{\partial y}\bigg)dy=0$$
And then
$$\frac{\partial k}{\partial x}=1$$
$$\frac{\partial l}{\partial x}=2x+y+2z\frac{\partial z}{\partial x}$$
$$\frac{\partial k}{\partial y}=1$$
$$\frac{\partial l}{\partial y}=x+2z\frac{\partial z}{\partial y}$$
By replacing these equations into total differenatial
$$df=\bigg(\frac{\partial f}{\partial k}+\frac{\partial f}{\partial l}\bigg(2x+y+2z\frac{\partial z}{\partial x}\bigg)\bigg)dx+\bigg(\frac{\partial f}{\partial k}+\frac{\partial f}{\partial l}\bigg(x+2z\frac{\partial z}{\partial y}\bigg)\bigg)dy=0$$
Without loss of generality
$$\frac{\partial f}{\partial k}+\frac{\partial f}{\partial l}\bigg(2x+y+2z\frac{\partial z}{\partial x}\bigg)=0$$
$$\frac{\partial f}{\partial k}+\frac{\partial f}{\partial l}\bigg(x+2z\frac{\partial z}{\partial y}\bigg)=0$$
We can combine both equations such that
$$\frac{\partial f}{\partial k}+\frac{\partial f}{\partial l}\bigg(2x+y+2z\frac{\partial z}{\partial x}\bigg)=\frac{\partial f}{\partial k}+\frac{\partial f}{\partial l}\bigg(x+2z\frac{\partial z}{\partial y}\bigg)$$
By substracting $\frac{\partial f}{\partial k}$ and then dividing by $\frac{\partial f}{\partial l}$ we have
$$2x+y+2z\frac{\partial z}{\partial x}=x+2z\frac{\partial z}{\partial y}$$
which can be simplified to
$$x+y=2z\bigg(\frac{\partial z}{\partial y}-\frac{\partial z}{\partial x}\bigg)$$
