How can we show that the functions are differentiable? Show that the following functions $$f(x, y)=\frac{xy}{\sqrt{x^2+y^2}} \\ f(x, y)=\frac{x^2y}{x^4+y^2}$$ are differentiable at each point of the domain. Determine which of them is $C^1$. 
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The domain of the functions is $\mathbb{R}^2 \setminus \{(0,0)\}$. 
How can we show that the functions are differentiable at each point of the domain?? 
Do we have to find the partial derivatives?? 
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EDIT: 
$$f(x, y)=\frac{xy}{\sqrt{x^2+y^2}}: \\ \frac{\partial{f}}{\partial{x}}=\frac{y\sqrt{x^2+y^2}-xy\frac{1}{2\sqrt{x^2+y^2}}2x}{x^2+y^2}=\frac{y(x^2+y^2)-x^2y}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{y^3}{\sqrt{x^2+y^2}(x^2+y^2)} \\ \frac{\partial{f}}{\partial{y}}=\frac{x\sqrt{x^2+y^2}-xy\frac{1}{2\sqrt{x^2+y^2}}2y}{x^2+y^2}=\frac{x(x^2+y^2)-xy^2}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{x^3}{\sqrt{x^2+y^2}(x^2+y^2)}$$ 
Since the partial derivatives are continuous at $\mathbb{R}^2 \setminus \{ (0,0)\}$, $f$ is differentiable. 
Is this correct?? 
Could I improve something at the fomrmulation?? 

Since $(0,0)$ isn't in the domain, then we don't need to worry about that point. The necessary
 and sufficient conditions for multi-variable differentiation are the existence of all the partials, and their continuity at a given point.
