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May i know if my approach is correct?

We need to solve the Cauchy Riemann equations to deduce which points $f(z)$ is differentiable at. Let $u(x,y) = x^{2}-y, v(x,y)= xy^{2}$ and thus $$u_x = 2x ~,~ u_y = -1~,~v_x =y^{2}~,~v_y = 2xy$$

we clearly see that $u_x,u_y,v_x,v_y$ are continuous on $\mathbb{C}$ since they are polynomials, thus to find where $f$ is differentiable, we proceed to solve the Cauchy - Riemann equations. $$u_x = v_y, u_y = -v_x$$

\begin{equation} u_x = v_y, u_y = -v_x \Rightarrow \begin{cases} 2x=2xy \\ y^{2} = 1 \end{cases} \end{equation}

By solving the above equation, we have two sets of solutions where $(x,1),(0,-1)$ and hence $f$ is differentiable at $z = -i$ and $z= x+i$ for all $x \in \mathbb{R}$

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    Why you omit $(0,1)$ from your solutions.? – 2017-01-30
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    @MyGlasses he did not, it is a part of $(x,1)$ – 2017-01-30
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    Ok. very good. .. – 2017-01-30
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    i think this is correct – 2017-01-30

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