Prove that $f$ is continuous in $(0,0)$ and check if $f$ differentiable in $(0,0)$ and if $f$ is uniformly continuous on $\mathbb R^2$ 
Let $f: \mathbb R^2 \to \mathbb R$ such that:
  $$f(x,y)= \begin{cases} \frac{(x+y)^3}{\sqrt{1+x^2+y^2}-1}, (x,y)\neq (0,0) \\ 0, (x,y)=(0,0) \end{cases}$$
  a) Prove that $f$ is continuous in $(0,0)$.
  b) Is $f$ differentiable in $(0,0)$?
  c) Is $f$ is uniformly continuous on $\mathbb R^2$?

My try:
a) $f(0,0)=0$ so to prove that $f$ is continuous in $(0,0)$ we should show that $\lim _{(x,y)\to (0,0)} \frac{(x+y)^3}{\sqrt{1+x^2+y^2}-1} =0$. So let's take a $(x_n,y_n)$ such that $(x_n,y_n)\to (0,0)$:
$$
\frac{(x_n+y_n)^3}{\sqrt{1+x_n^2+y_n^2}-1} 
\le \frac{(x_n^2+y_n^2)^3}{\sqrt{1+x_n^2+y_n^2}-1}\\
=\frac{(x_n^2+y_n^2)^3(\sqrt{1+x_n^2+y_n^2}+1)}{x_n^2+y_n^2}\\
=(x_n^2+y_n^2)^2(\sqrt{1+x_n^2+y_n^2}+1) \to (0+0)(\sqrt{1+0+0}-1)=0
$$
$$
\frac{(x_n+y_n)^3}{\sqrt{1+x_n^2+y_n^2}-1} \ge \frac{(x_n+y_n)^3}{\sqrt{1+x_n^2+y_n^2}}\ge \frac{(x_n+y_n)}{\sqrt{1+x_n^2+y_n^2}} \ge \frac{(x_n+y_n)}{1+x_n^2+y_n^2} \to \frac{0+0}{1+0+0}=0
$$
So from theorem of three functions $\lim _{(x,y)\to (0,0)} \frac{(x+y)^3}{\sqrt{1+x^2+y^2}-1} =0$.
b) We already know that $f$ is continuous in $(0,0)$ so to prove that $f$ is differentiable in $(0,0)$ we should still show that exist linear map $A: \mathbb R^n \to \mathbb R^m$ such that: $$\lim_{||h||\to 0} \frac{||f(a+h)-f(a)-Ah||}{||h||}=0$$So: 
$$
\lim_{||h||\to 0} \frac{||f(a+h)-f(a)-Ah||}{||h||}
=\lim_{||h||\to 0} \frac{|| \frac{(h_1+h_2)^3}{\sqrt{1+h_1^2+h_2^2}-1}-Ah||}{\sqrt{h_1^2+h_2^2}}\\
=\lim_{||h||\to 0} ||\frac{(h_1+h_2)^3(\sqrt{1+h_1^2+h_2^2}+1)-Ah(h_1^2+h_2^2)}{(h_1^2+h_2^2)\sqrt{h_1^2+h_2^2}}||\\
=\lim_{||h||\to 0} ||\frac{(h_1+h_2)^3(\sqrt{1+h_1^2+h_2^2}+1)-Ah(h_1^2+h_2^2)}{(h_1^2+h_2^2)^{\frac 32}}||
$$
My questions:
 In a) I would like someone to check if my solution is correct
 In b) I don't know what I can do the next

Hints:
a) No, $(x+y)^3\le (x^2+y^2)^3$ is false. Take $x=1/2,y=0$ to see this. But this is true:
$$|f(x,y)| \le \frac{(|x|+|y|)^3}{\sqrt {1+x^2+y^2}-1}=\frac{(|x|+|y|)^3}{x^2+y^2}(\sqrt {1+x^2+y^2}+1).$$
b) If $f$ is differentiable at $(0,0),$ then
$$f(x,y) = \nabla f(0,0)\cdot (x,y) + o\left(\sqrt {x^2+y^2}\right).$$
I'm getting $\nabla f(0,0)=(2,2).$
c) If $f$ were uniformly continuous, then $x\to f(x,0)$ would be uniformly continuous on $\mathbb R.$ And a uniformly continuous function on $\mathbb R$ is always bounded above by a linear function. But $f(x,0)=x(\sqrt {1+x^2}+1).$
