I come from this post but I think that no one was going to answer me again since I asked several questions on the same subject. Apologies for the untidiness.
To study the differentiability of $$f(x,y)=\begin{cases}x^2+y,\ x\geq 1\\3x-y,\ x<1\end{cases}$$ at $(1,1)$ we have to find the partial derivatives of $f$ at that point. So we have:
$$f'_x(1,1)=f((1,1);(1,0))=\lim_{h\to 0}{\frac{f(1+h,1)-f(1,1)}{h}}=\lim_{h\to 0}{\frac{f(1+h,1)-2}{h}},$$
but at the evaluated point we can approach both from the left ($ h \to 0^-$) and from the right ($ h \to 0^+ $) since it is a piecewise function. So
$$\displaystyle\lim_{h\to 0}{\frac{f(1+h,1)-2}{h}}=\begin{cases}\displaystyle\lim_{h\to 0^+}{\frac{f(1+h,1)-2}{h}}&=&\displaystyle\lim_{h\to 0^+}{\frac{{(1+h)}^2+1-2}{h}}=\displaystyle\lim_{h\to 0^+}{\frac{h^2+2h}{h}}&=&2\\ \displaystyle\lim_{h\to 0^-}{\frac{f(1+h,1)-2}{h}}&=&\displaystyle\lim_{h\to 0^-}{\frac{3(1+h)-1-2}{h}}=\displaystyle\lim_{h\to 0^-}{\frac{3h}{h}}&=&3\end{cases}$$
but $$\displaystyle\lim_{h\to 0^+}{\frac{f(1+h,1)-2}{h}}\neq \displaystyle\lim_{h\to 0^-}{\frac{f(1+h,1)-2}{h}}$$ thus $$\not\exists f'_x(1,1),$$ therefore (since one of the conditions for a function to be differentiable at a point is that there are partial derivatives) $$\boxed{f\quad\text{is not differentiable at}\quad (1,1)}.$$
Is that correct?
Thank you in advance!