Showing $g:\mathbb{R}^2\rightarrow\mathbb{R}$ is continuously differentiable. Given
$$g(x,y)=\bigg\{\frac{x^2y^4}{(x^2+y^2)} \text{ if }(x,y)\neq(0,0)$$
$$g(0,0)=0$$
My attempt:
$\frac{\partial g}{\partial x}=\frac{2xy^6}{(x^2+y^2)^2},\frac{\partial g}{\partial y}=\frac{x^2(4x^2y^3+2y^5)}{(x^2+y^2)^2}$ in the neighborhood around $(0,0)$
$$\therefore \frac{\partial g}{\partial x}(0,0)=\lim_{t\rightarrow 0} \frac{f(t,0)-f(0,0)}{t}=\frac{t^2(0)}{t^4}\cdot\frac{1}{t}=0$$
$$\therefore \frac{\partial g}{\partial y}(0,0)=\lim_{t\rightarrow 0} \frac{f(0,t)-f(0,0)}{t}=\frac{0(0+2t^5)}{t^4}\cdot\frac{1}{t}=0$$
Since $g_x,g_y$ is continuous.

You have to prove that $\frac {\partial g} {\partial x} (x,y) \to 0$ as $(x,y) \to 0$ and $\frac {\partial g} {\partial y} (x,y) \to 0$ as $(x,y) \to 0$. For this use the following: $2|x||y| \leq x^{2}+y^{2}$ and $y^{2} \leq x^{2}+y^{2}$; from these we get $|\frac {\partial g} {\partial x} (x,y)| \leq y^{4} \to 0$. Similarly, use $x^{2} \leq x^{2}+y^{2}$ and $y^{2} \leq x^{2}+y^{2}$ for the partial derivative w.r.t. $y$. 
