Prove $\frac{\partial u}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial v}{\partial x}\frac{\partial x}{\partial v}=1$ If $$u=f(x,y)$$
and
$$v=g(x,y)$$
are differentiable.

Prove $$\frac{\partial u}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial v}{\partial x}\frac{\partial x}{\partial v}=1$$

I knew that for one variable case 
$$\frac{du}{dx}=\frac{1}{\frac{dx}{du}}$$
 but I don't know how to do it in two variables. I also tried to differentiate both equations regarding to $x$, still no clue about how to solve it...Any help? Thank you~
The answer by @W.mu is a poorly formulated, but correct.

Assume that the map
$$
(x,y) \mapsto (f(x,y),g(x,y))
$$
is invertible. That is there exist functions $d,e$ such that
$$
(u,v) \mapsto (d(u,v),e(u,v))
$$
is the inverse of the previous map, that is, giving $(x,y)$ as function of $(u,v)$. Thus, we have
$$
(x,y) \mapsto \Bigl(d\bigl(f(x,y),g(x,y)\bigr),e\bigl(f(x,y),g(x,y)\bigr)\Bigr) = (x,y).
$$
Then you can differentiate the first component of this composition with respect to $x$ to get
$$
  \frac{\partial d}{\partial u}\frac{\partial f}{\partial x}
 +\frac{\partial d}{\partial v}\frac{\partial g}{\partial x}
= \frac{\partial x}{\partial u}\frac{\partial u}{\partial x}
 +\frac{\partial x}{\partial v}\frac{\partial v}{\partial x} = 1.
$$
In the last step I have been sloppy in interpreting $x = x(u,v) = d(u,v,)$, etc. This is fairly common in mathematics writing, unfortunately, since otherwise you quickly run out of symbols to denote every new function, and the notation becomes very heavy.
If we can suppose the transformation is invertible.
From IFT, we have:
$$x=x(u,v);y=y(u,v)$$
then we differentiate the first one to x:
$$1=x_uu_x+x_vv_x$$
Isn't it?