Is $f$ differentiable at $(x,y)$? I am working on a practice question, and I am not sure if what I have done would be considered, 'complete justification'. I would greatly appreciate some feedback or helpful advice on how it could be better etc. The question is here:
Let $f: \mathbb {R}^2 \to \mathbb{R} $ be a function defined by:
$$ \
 f(x,y) =
  \begin{cases}
   \frac{\sin(x^2 + y^2)}{x^2 + y^2} & \text{if } (x,y) \ne (0,0) \\
   1       & \text{if } (x,y) = (0,0)
  \end{cases}
$$
Is $f$ diff'ble at $(x,y) = (0,0)$? 
Here is what I have:
By definition, $f(x,y)$ is diff'ble at $(0,0)$ if
$$ \lim_{(x,y) \to (0,0)} f(x,y)= \frac{f(x,y) - [f(0,0) + f_{x}(0,0)(x-0) + f_{y}(0,0)(y-0)]}{\sqrt{x^2 + y^2}} = 0\tag{*}$$
Since $f(x,y)$ is piecewise, $f_{x}(0,0)$ and $f_{y}(0,0)$ is derived from 1st principles:
So, $$\begin{align}f_{x}(0,0) &= \lim_{h \to 0}  \frac{\frac{\sin((0+h)^2 + (0)^2)}{(0+h)^2 + (0)^2} - f_{x}(0,0)}{h} \\
&=  \lim_{h \to 0} \frac{\frac{\sin(h^2)}{h^2} - 1}{h}\\
&=  \lim_{h \to 0}  \frac{\sin(h^2) - h^2}{h^3} \\
&=  \lim_{h \to 0}  \frac{2h \cos(h^2) - 2h}{3h^2}\end{align} $$ by L'Hopital's rule, apply this twice more I can see that the limit is $0$. Similarly, $f_y(0,0)$ is derived the same way.
Then from $(*)$, I have:
$$f_{x}(0,0) = \lim_{h \to 0}  \frac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}} $$
From here I haven't had luck trying to get this to $0$. So instead, I try to make it easier.
Since $f_x(0,0)$ and $f_y(0,0)$ exist we must evaluate whether $f_x(x,y)$ is continuous at $(x,y)=(0,0)$ i.e. if
$$\lim_{(x,y) \to (0,0)}  \frac{\sin(x^2 + y^2)}{x^2 + y^2} = 1 $$ then  $f(x,y)$ is differentiable at $(0,0)$. So letting $u = x^2 + y^2$,
$$\lim_{(x,y) \to (0,0)}  \frac{\sin(x^2 + y^2)}{x^2 + y^2} = \lim_{ (x,y) \to (0,0)}  \frac{\sin(u)}{u} = 1 $$ by L'Hopital once more.
Therefore, $f$ is differentiable at (0,0).
I feel like I am perhaps not completely justifying this, because I take a short cut - but is it valid? Should I perhaps use a epsilon-delta proof? Or is there a way to work with that tricky limit I have in $(*)$?
Many thanks for the help in advance!

You've already found
$$\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{f(x,y) - [f(0,0) + f_{x}(0,0)(x-0) + f_{y}(0,0)(y-0)]}{\sqrt{x^2 + y^2}}\right)=\lim \limits_{(x,y)\to (0,0)}\left( \dfrac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}}\right).$$
To evaluate $\lim \limits_{(x,y)\to (0,0)}\left( \dfrac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}}\right)$ simplify the denominator and change the variables with $x=\rho \cos(\theta)$ and $y=\rho\sin(\theta)$ to get
$$\begin{align}\lim \limits_{(x,y)\to (0,0)}\left( \dfrac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}}\right)&=\lim \limits_{(x,y) \to (0,0)}\left(\dfrac{\sin(x^2+y^2)-(x^2+y^2)}{(x^2+y^2)^{3/2}}\right)\\
&=\lim \limits_{\rho \to 0}\left(\dfrac{\sin\left(\rho^2\right)-\rho ^2}{|\rho| ^3}\right)\\
&=\lim \limits_{\rho \to 0}\left(\dfrac{\rho ^2+o\left(\rho^2\right)-\rho^2}{|\rho|^3}\right)\\
&=0.\end{align}$$
