$f,g$ diffirentiable function at point $(x_0, y_0)$ how to show that $fg$ diffirentiable function at point $(x_0, y_0)$? I guess there is pretty simple way of showing the statement below.. I tried using definition but it seem complicated. 

Suppose $f, g: \Bbb R^{2} \to \Bbb R$.  Prove if $f, g$ are differentiable at $(x_{0}, y_{0})$, then the product, $fg$, is differentiable at $(x_{0}, y_{0})$.

 A: For all $x$ and $h$ such that $x$ and $x+h$ are in the domain of differentiability of $f$ and $g$, we have
$$(fg)(x+h)=f(x+h)g(x+h)=\left(f(x)+\mathrm{d}f(x)(h)+o(h)\right)\left(g(x)+\mathrm{d}g(x)(h)+o(h)\right)$$
$$=f(x)g(x)+f(x)\mathrm{d}g(x)(h)+g(x)\mathrm{d}f(x)(h)+o(h)$$
and thus $fg$ is differentiable at $x$ and we have
$$\mathrm{d}(fg)(x)(h)=f(x)\mathrm{d}g(x)(h)+g(x)\mathrm{d}f(x)(h).$$
EDIT: for your purpose, $x=(x_0,y_0)\in\mathbb{R}^2$ and thus $h=(h_0,k_0)\in\mathbb{R}^2$ too.
A: $fg=\frac {(f+g)^2-(f-g)^2}{4}$
A: It may simply your life a bit to prove that  $P(x,y) = xy$ is a differentiable function from $\mathbb{R}^2 \to \mathbb{R}$.  Also prove that the function $Q: \mathbb{R}^2 
\to \mathbb{R}^2$ given by $Q(x,y) = (f(x,y),g(x,y))$ is differentiable at $(x_0,y_0)$.  The function you are interested in is $P \circ Q$, which is differentiable by the chain rule.
A: Since you are writing the product, I assume that both functions take value in $\mathbb{R}$. 
What is the equivalent result of yours in 1 dimension? How would you prove it?
Hint:
In 1D : $(fg)' = f'g + fg'$
By definition of $\text{d}f$:
$f(x + h) = f(x) + \text{d}f(x).h + o(\lvert h \rvert)$
A: We know that $f'(x,y):\mathbb R^{2}\rightarrow \mathbb R$ is defined by 
$(h_{1},h_{2}) \mapsto h_{1}\frac{\partial f}{\partial x}(x,y) +h_{2}\frac{\partial f}{\partial y}(x,y)$ or $f'(x,y)\cdot \textbf h \equiv  {\triangledown }f(x,y)\cdot \textbf h $ 
and similarly for $g'(x,y)$.
Now fix $\textbf x =(x,y)$ and define $f_{1}=f-f(\textbf x)$; $g_{1}=g-g(\textbf x)$ so that 
$\tag 1 f_{1}(\textbf x)=g_{1}(\textbf x)=0$
We also have
$\tag 2 fg=f_{1}g_{1}+g(\textbf x)f_{1}+f(\textbf x)g_{1}+f(\textbf x)g(\textbf x)$
And we know that there is a $K>0$ such that
$\tag 3 \vert f(\textbf x+\textbf y)-f(\textbf y)\vert \leq K\vert \textbf y\vert$
which follows directly from the definition of derivative. A similar result holds for $g$. 
Now $\triangledown (f_{1}g_{1})(\textbf x):\mathbb R^{2}\rightarrow \mathbb R$ is the unique linear transformation that satisfies
$\left | \frac{f_{1}(\textbf x+\textbf y)(g_{1}(\textbf x+\textbf y)-f_{1}(\textbf x)g_{1}(\textbf x)-\triangledown (f_{1}g_{1})(\textbf x)\cdot \textbf y}{\textbf y} \right |\rightarrow 0$ as $\vert y\vert \rightarrow 0$.
But now, using $(1)$ we get 
$\tag 4 \left | \frac{f_{1}(\textbf x+\textbf y)(g_{1}(\textbf x+\textbf y)-\triangledown (f_{1}g_{1})(\textbf x)\cdot \textbf y}{\textbf y} \right |\rightarrow 0$ as $\vert y\vert \rightarrow 0$.
Now, observe that $f_{1}(\textbf x+\textbf y)g_{1}(\textbf x+\textbf y)=(f(\textbf x+\textbf y)-f(\textbf x))(g(\textbf x+\textbf y)-g(\textbf x))$ so we can also apply $(3)$ to write
$\tag 5 \vert (f_{1}(\textbf x+\textbf y)(g_{1}(\textbf x+\textbf y)\vert \leq C\vert \textbf y\vert ^{2}$
Finally, combining $(4)$ and $(5)$ we may conclude that $\triangledown (f_{1}g_{1})(\textbf x)=0$ and now inspection of $(2)$ gives the result. 
