$f(z)$ has essential singularity at $z=0$, what type of singularity $f(z^2+z)$ has?
$f(z)$ has essential singularity at $z=0$ so it can be written has $\sum_{n=0}^{-\infty} c_nz^n=c_0+\frac{c_{-1}}{z}+\frac{c_{-2}}{z^2}+...+$ substituting $z=z^2+z$ will leave all the power to be negative e.g $c_0+\frac{c_{-1}}{z^2+z}+\frac{c_{-2}}{(z^2+z)^2}+...+$
So $f(z^2+z)$ has essential singularity too, moreover can we even "get rid" of essential singularity?
Here is a rigorous proof of the fact that $g(z)\equiv f(z^{2}+z)$ has an essential singularity without using Laurent series. If $g$ has a removable singularity at $0$ then so it is bounded in a neighborhood of $0$, say $|g(z)| <M$ for $|z|<\delta $ with $0<\delta <1$. If $|w|$ is sufficiently small take $z=\frac {-1+c} 2$ where $c$ is such that $c^{2}=1+4z$ and $\Re c>0$. Then $z^{2}+z=w$ so we get $|f(w)|<M$. Thus $f$ is bounded in a neighborhood of $0$ contradicting the fact that $f$ has an essential singularity. If $g$ has a pole at $0$ then $|g(z)| \to \infty $ as $z \to 0$ and an argument similar to the previous one shows that $|f(z)| \to \infty $ as $z \to 0$, again a contradiction.