Calculation of a partial derivative I have to find the $\frac{\partial}{\partial x}\left( f(x,y)\right)=\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$.
1) Thinking of that as $\frac{\partial}{\partial x}\left( (x^2+y^2)^{-\frac{1}{2}}\right)$ and $\frac{\partial}{\partial a}x^a = ax^{a-1}$ I get $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right) = -\frac{x}{\Vert (x,y) \Vert^3}$.
2) Thinking of that as $\frac{\partial}{\partial x}\left( (\sqrt{x^2+y^2})^{-1}\right)$ and $\frac{\partial}{\partial a}\sqrt{a}=\frac{1}{2\sqrt{a}}$ I get $x \cdot (x^2+y^2)$.
3) Thinking of that as $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$ and $\frac{\partial}{\partial a}\frac{1}{a}=-\frac{1}{a^2}$ I get $-\frac{2x}{x^2+y^2}$.
Why 2) and 3) are not correct?

Note:
$$\frac{\partial}{\partial x}\left( \left(\sqrt{x^2+y^2}\right)^{-1}\right)=(-1)\left(\sqrt{x^2+y^2}\right)^{-2}\frac{\partial}{\partial x}\sqrt{x^2+y^2}$$
$$\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)=\frac{0-1\cdot \frac{\partial}{\partial x}\sqrt{x^2+y^2}}{\left(\sqrt{x^2+y^2}\right)^2}$$
