Clarifications about the correct way to solve exercises (continuity, partial derivatives, differentiability) I need some clarifications about the correct way to solve an exercise.
I have this function:
$$f(x,y)=\frac{(x-1)y^2}{\sin^2\sqrt{(x-1)^2+y^2}}$$
and I have to 


*

*analyse the existence of partial derivatives and the differentiability in (1,0)

*say if f is differentiable in (0,0)
About point 1:
I have this doubt. Do I have to calculate the partial derivatives considering the expression of g(x,y) in (1,0) (i.e. g(1,0)=0 -> $\displaystyle {\partial f}/\partial x=\partial 0 / \partial x=0={\partial f}/\partial y $ (*) or do I have to use 
$\displaystyle\frac{\partial}{\partial x}g(1,0)=\lim_{t->0} \frac{f(1+t,0)-g(1,0)}{t}$ ? (**) Or do I have to find the expressions of the partial derivatives without using the limit and then do the limit for (x,y)->(1,0)?
If I want to analyse if $\frac{\partial}{\partial x}g(1,0)$ is continue, do I have to compare (*) and (**)?
About point 2:
If a function has continue partial derivatives in a point, it is differentiable in that point.
So I have thought that in Dom f, the partial derivatives are continue because thay're composition of continue function. Since $0 \in Dom f$, g(x,y) is differentiable in (0,0). 
Is this reasoning ok, or do I have to check with calculation?
Many thanks for your help!
 A: 1) If you take polar coordinates you get immediatly $$\lim_{\rho\rightarrow0^{+}}\frac{\rho^{2}\sin^{2}\left(\theta\right)\left(\rho\cos\left(\theta\right)-1\right)}{\sin^{2}\left(\sqrt{\rho^{2}-2\rho\cos\theta+1}\right)}=0.
 $$ 2) See Dr. MV answer.
3) A function is differentiable in a point $\left(x_{0},y_{0}\right)
 $ if the partials derivative exist in that point and if $$\lim_{\left(h,k\right)\rightarrow\left(0,0\right)}\frac{f\left(x_{0}+h,y_{0}+k\right)-f\left(x_{0},y_{0}\right)-\frac{\partial f}{\partial x}\left(x_{0},y_{0}\right)h-\frac{\partial f}{\partial y}\left(x_{0},y_{0}\right)k}{\sqrt{h^{2}+k^{2}}}=0.
 $$
A: As shown by Marco Cantarin, we have that $f$ has a removable discontinuity at $(1,0)$ and we may define a new function, say, $\bar f$, that is equal to $f$ for $(x,y)\ne (1,0)$ and equal to $1$ for $(x,y)=(1,0)$. 
Now, let's analyze the partial derivatives at $(1,0)$.  To that end, we use the definition of the first partial derivatives to expose that 
$$\bar f_x(1,0)=\lim_{h\to 1}\frac{\bar f(1+h,0)-\bar f(1,0)}{h}=\lim_{h\to 1}\frac{0-0}{h}=0$$
$$\bar f_y(1,0)=\lim_{h\to 0}\frac{\bar f(1,h)-\bar f(1,0)}{h}=\lim_{h\to 1}\frac{0-0}{h}=0$$
To show differentiability at $(1,0)$ we see that 
$$\begin{align}
\bar f'(1,0) &\equiv \lim_{(h,k)\to (1,0)}\frac{\bar f(1+h,k)-\bar f(1,0)-\bar f_x(1,0)h-\bar f_y(1,0)k}{\sqrt{h^2+k^2}}\\\\
&=\lim_{(h,k)\to (1,0)}\frac{\frac{hk^2}{\sin \sqrt{h^2+k^2}}-0-0\cdot h-0\cdot k}{\sqrt{h^2+k^2}}\\\\
&=\lim_{(h,k)\to (1,0)}\frac{hk^2}{h^2+k^2}\cdot \lim_{(h,k)\to (1,0)}\frac{\sqrt{h^2+k^2}}{\sin \sqrt{h^2+k^2}}\\\\
&=\lim_{(h,k)\to (1,0)}\frac{hk^2}{h^2+k^2}\\\\
&=0
\end{align}$$
since 
$$\frac{hk^2}{h^2+k^2}\le \frac{h(h^2+k^2)}{h^2+k^2}=h$$
Thus, $f$ is indeed differentiable at $(1,0)$!
