Cauchy Riemann Equations for $h(z)=|z|^2z^2$ A past exam question I have come across is...
Is the following function differentiable on $\mathbb{C}$
$h(z)=|z|^2z^2$
I have calculated $u(x,y)=x^4-y^4$ and $v(x,y)=2x^3y+2xy^3$
I then have
$\frac{du}{dx}=4x^3,$ 
$\frac{dv}{dx}=6x^2y+2y^3$
$ \frac{dv}{dy}=-4y^3$
$ \frac{dv}{dy}=2x^3+6xy^2$
hence these do not satisfy the Cauchy Riemann equations as $\frac{du}{dx} \neq\frac{dv}{dy}, -\frac{dv}{dx} \neq \frac{du}{dy}$
Is this enough to show that its not differentiable on $\mathbb{C}$??

Yes. Though a simpler derivation (though I am not sure whether your course has covered the notation) is the following. 
Writing $z = x+iy$ and $\bar{z} = x - iy$. Formally $z$ and $\bar{z}$ form a coordinate system of $\mathbb{C}$ (in the sense that $(x,y)$ has two component and $(z,\bar{z})$ has two components, this can be made more precise in the context of complex geometry, but just take this as a given for now) by pretending that $z$ and $\bar{z}$ are "real" numbers. The usual change of variables formula allows one to write $\partial_x, \partial_y$ in terms of $\partial_z$ and $\partial_{\bar{z}}$. 
Now here's the kicker: the Cauchy-Riemann equations can be written as 
$$ \frac{\partial f}{\partial \bar{z}} = 0 $$
in this coordinate system. 
Your function $h$ can be written as
$$ h(z,\bar{z}) = z^3 \bar{z} $$
and hence
$$ \frac{\partial h}{\partial \bar{z}} = z^3 \neq 0$$ 
when $z \neq 0$. This shows that your function does not satisfy the Cauchy-Riemann equations and so that it is not holomorphic. 
