Let $f(x+iy)=x^2-y^2 + 5xi$. So hence $u(x,y)=x^2-y^2$ and $v(x,y)=5x$
In my notes it calculated $\frac{\partial u}{\partial x}$ at $0$ as follows:
$\frac{\partial u}{\partial x}(0,0)=\displaystyle\lim_{h\rightarrow 0}\frac{u(x+h,y)-u(x,y)}{h} \\=\displaystyle\lim_{h\rightarrow 0}\frac{u(h,0)-u(0,0)}{h}\\=\displaystyle\lim_{h\rightarrow 0}\frac{h^2}{h}=\displaystyle\lim_{h\rightarrow 0}h=0$
But is it possible to calculate $\frac{\partial u}{\partial x}$ at $0$ by just finding that $\frac{\partial u}{\partial x} = 2x$, and then substituting $x=0,y=0$ and thus getting $\frac{\partial u}{\partial x}=0$?
If so, it seems easier that way rather than taking limits as above.