Prove that $f$ is differentiable in $(0,0)$ if and only if $\lim_{t\to0+} g(t)$ exists Let $g:[0,\infty)\to\mathbb{R}$ be a mapping and $f(x,y)=xg(\sqrt{x^2+y^2})$ for all $(x,y)\in\mathbb{R^2}$.
Prove that $f$ is differentiable in $(0,0)$ $\iff$ $\lim_{t\to0+} g(t)$ exists.
My attempt:
$\implies:$ We suppose that $f$ is differentiable in $(0,0)$, that is, there is a linear mapping $A=(a_1 \,\, a_2)$ such that $$f(x,y)-f(0,0)-A(x,y) = R(x,y)$$ where R is a remainder term satisfying $\lim_{(x,y)\to(0,0)}\frac{R(x,y)}{|(x,y)|} = 0$.
Given the definition of $f$, we see that $f(0,0)=0$, so that $$f(x,y)-A(x,y) = R(x,y) \\ \iff xg\big(\sqrt{x^2+y^2}\big)-A(x,y) = R(x,y) \\ \iff g\big(\sqrt{x^2+y^2}\big) = \frac{R(x,y)}{x}+\frac{1}{x}A(x,y) = \frac{R(x,y)}{x}+\frac{1}{x}(a_1x + a_2y).$$
Now we use polar coordinates: $x=r\cos\theta$ and $y=r\sin\theta$, which gives $$g(r) = \frac{R(r\cos\theta, r\sin\theta)}{r\cos\theta}+\frac{1}{r\cos\theta}(a_1r\cos\theta+a_2r\sin\theta) \\= \frac{R(r\cos\theta, r\sin\theta)}{r\cos\theta}+a_1+a_2\frac{\sin\theta}{\cos\theta}$$
Now taking the limit as $r\to0$ the first term vanishes and we see that the limit is $$a_1+a_2\frac{\sin\theta}{\cos\theta}.$$
$\impliedby:$ Now we suppose that $\lim_{t\to0+} g(t)$ exists. I will calculate the two partial derivatives and then show that the matrix $$A = \big(\partial_1 f(0,0) \,\,\, \partial_2 f(0,0)\big)$$ is indeed the derivative of $f$ in $(0,0)$.
We have that $$\partial_1 f(0,0) = \lim_{t\to0} \frac{f(t,0)}{t} = \lim_{t\to0}\frac{tg(t)}{t}=\lim_{t\to0}g(t) = M$$ for some finite $M$; and $$\partial_2 f(0,0)=\lim_{t\to0}\frac{f(0,t)}{t} = 0.$$
We shall now investigate $$\lim_{(x,y)\to(0,0)} \frac{f(x,y) -f(0,0) - A(x,y)}{\sqrt{x^2+y^2}}.$$ Using polar coordinates this becomes $$\lim_{r\to0} \frac{r\cos\theta \,g(r) - Mr\cos\theta}{r} = \lim_{r\to0} \cos\theta \,g(r) - M\cos\theta = 0,$$ as desired.
This post turned out much longer than I had anticipated -- I'm sure I made it far more complicated than necessary :)
 A: For the first implication, if $f$ is differentiable at $(0,0)$, then in particular it admits a partial derivative with respect to $x$ at $0$, so that the following limit exists : 
$$
\frac{\partial f}{\partial x}(0,0) = \lim_{x \to 0^+} \frac{f(x,0) - f(0,0)}{\sqrt{x^2+0^2}} = \lim_{x \to 0^+} \frac{x g\left( \sqrt{x^2+0^2} \right) - 0}{x} = \lim_{x \to 0^+} g(x).
$$
Conversely, define $M \overset{def}= \lim_{x \to 0^+} g(x)$. Then 
$$
\lim_{\sqrt{x^2+y^2} \to 0} \frac{f(x,y) - Mx}{\sqrt{x^2+y^2}} = \lim_{\sqrt{x^2+y^2}} \frac{x g \left( \sqrt{x^2+y^2} \right) - Mx}{\sqrt{x^2+y^2}} = \lim_{\sqrt{x^2+y^2} \to 0} \left( \frac{x}{\sqrt{x^2+y^2}} \right) \left( g \! \left( \sqrt{x^2+y^2} \right) - M \right) = 0.
$$
The last limit is equal to zero because 
$$
\left| \left( \frac{x}{\sqrt{x^2+y^2}} \right) \left( g \!\left( \sqrt{x^2+y^2} \right) - M \right) \right| \le \left| \, g \! \left( \sqrt{x^2+y^2} \right) - M \right| \to 0.
$$
This proves $f$ is differentiable (with derivative $A(x,y) = Mx$, as expected).
You were almost there! Honestly, I traced on your steps. You did a good job =)
Hope that helps,
