Classifying singular points of $\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}$ 
I am trying to classify the singular points of the function $$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}.$$

My attempt:
$$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}=\frac{\sin(z^2)}{z^2\left(z-\frac{\pi}{4}\right)}.$$
Hence the singular points are $z=0,\frac{\pi}{4}$. For classification:
$z=\frac{\pi}{4}$ is a simple pole as $\left(z-\frac{\pi}{4}\right)$ is a simple zero.
For $z=0$, we can see that $z^2\left(z-\frac{\pi}{4}\right)$ is a zero of order $2$. Also,$\ \sin(z^2)$ is a zero of order $2$ for $z=0$. Hence, $z=0$ is a removable singularity. 
I am unsure about the $z=0$ case, particularly if $\sin(z^2)$ is a zero of order $2$ for $z=0$.

It is correct.
Also you can notice that
$$\lim_{z\to 0} \frac{\sin(z^2)}{z^2\left(z-\frac{\pi}{4}\right)}= \lim_{z\to 0} \frac{\sin(z^2)}{z^2} \cdot \lim_{z\to 0}\frac1{z-\frac\pi4} = 1 \cdot \frac1{-\frac\pi4} = -\frac4\pi$$
so $z = 0$ is a removable singularity.
