Find the radius of convergence for the Taylor series of $f(z)$ at $z=0$. Let , $$f(z)=\frac{z}{\sin (z^2+z)}.$$
$(1)$ Define $f(0)$ so that $f$ becomes analytic  at $z=0$.
$(2)$ Find the radius of convergence for the Taylor series of $f(z)$ at $z=0$.

First part is done. If we define $f(0)=1$ then $f$ becomes analytic.
Now Taylor series expansion is $f(z)=\sum_{n=0}^{\infty}a_nz^n$ , where , $a_n=\frac{f^{(n)}(a)}{n!}$. From here how we find $a_n$ ?
$$\lim_{x\to 0}\frac{x}{\sin(x+x^2)}=\lim_{x\to 0}\frac{x}{x+x^2}\cdot\frac{x+x^2}{\sin(x+x^2)}=1$$
and the radius of convergence of the Taylor series in $x=0$ is given by the distance from the origin of the closest singularity. $\sin(z)$ vanishes only for $z\in \pi\mathbb{Z}$, hence:

$$ \rho = \frac{-1+\sqrt{1+4\pi}}{2}=1.34162771851\ldots$$
Hint: Let us firstly compute $f'(0).$ We have $f'(0)=\lim_{z\to 0} \frac{f(z)-f(0)}{z}=\lim_{z \to 0} \frac{z-\sin(z^2+z)}{z \sin(z^2+z)}.$ Since $\sin(z^2+z)=z^2+z+o(z)$ then $a_1=f'(0)=1.$
We have $$f'(z)=\frac{\sin(z^2+z)-z(2z+1)\cos(z^2+z)}{\sin^2(z^2+z)}.$$ Try to find recurrence formula to get the expression of $f^{(n)}(z)$ for each $z$ different from $0$ and take the limit after  that of $z$ to $0.$