Theorem: Suppose that $$f(z)=u(x,y)+i\,v(x,y)$$ and that $f’(z)$ exists at a point $z_0=x_0+i\,y_0$. Then the first order partial derivatives of $u$ and $v$ must exists at $(x_0,y_0)$, and they must satisfy the Cauchy-Riemann equations $$u_x=v_y,\quad u_y=-v_x$$ there. Also, $f’(z_0)$ can be written $$f’(z_0)=u_x+i\,v_x,$$ where these partial derivatives are to be evaluated at $(x_0,y_0)$.
I understand that these equations provide necessary conditions for the existence of $f’(z_0)$ which is weaker than sufficient conditions. So my book, in the examples, states that, if the Cauchy Riemann equations are not satisfied then the derivative a function does not exist. Therefore, my question is, would I be able to use this theorem to find points in which a function is not differentiable?
Yes. Take, for instance $f(z)=|z|^2$. In this case, $u(x,y)=x^2+y^2$ and $v\equiv0$. So,$$\left\{\begin{array}{l}u_x(x,y)=v_y(x,y)\\u_y(x,y)=-v_x(x,y)\end{array}\right.\iff\left\{\begin{array}{l}2x=0\\2y=0\end{array}\right.$$and therefore $f$ is not differentiable at any $z$ for which $z\neq0$. It turns out that $f$ is differentiable at $0$:$$f'(0)=\lim_{z\to0}\frac{f(z)-f(0)}{z}=\lim_{z\to0}\overline z=0.$$This can also be deduced from the Cauchy-Riemann equations: since the domain of $f$ is an open set and $u$ and $v$ are continuously differentiable there, $f$ is differentable at $x+yi$ if and only if $(x,y)$ is a solution of the Cauchy-Riemann equations.