Show identity with function of 2 variables. Let $f(x,y)$ - differentiable in $\mathbb{R}^2$.
Let $x=aX + bY$, $y=cX+dY$.
We put $g(X,Y) = f(aX+bY,cX+dY)$.
Show identity  : 
$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = X \frac{\partial g}{\partial X} + Y\frac{\partial g}{\partial Y}$
What should I do first? I've tried to count derivatives and see what happens but i got stuck.

Notice that
$$x=aX+bY$$
$$y=cX+dY$$
Given that
$$g(X,Y)=f(x,y)=f(aX+bY,cX+dY)$$
by the Chain Rule we obtain the result that is required.
The calculations wil we the following way:
$$\frac{\partial g}{\partial X} = \frac{\partial f}{\partial X} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial X} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial X} = a\frac{\partial f}{\partial x} + c\frac{\partial f}{\partial y}$$
and 
$$\frac{\partial g}{\partial Y} = \frac{\partial f}{\partial Y} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial Y} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial Y} = b\frac{\partial f}{\partial x} + d\frac{\partial f}{\partial y}$$
Consequently
$$X\frac{\partial g}{\partial X}+Y\frac{\partial g}{\partial Y}=(aX+bY)\frac{\partial f}{\partial x}+(cX+dY)\frac{\partial f}{\partial y}=x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}$$
I hope that all the calculations were clear enough.
