I have a doubt. In order to apply the theorem you have to meet the conditions : $f$ must be a meromorphic function that has only simple poles, is continuous, or has a removable singularity at 0.
I have tried to calculate the expansion at two points $0$ and $\pi$ of $\csc(z^2)$ which has double poles whereas if you see the following representation, both developments seem valid.
$$\sum _{k=2}^{\infty } \frac{2 e^{-i \pi k} x^2 \left(x^2+\pi \right)}{\pi \left(k^2-1\right) \left(\pi ^2 k^2-x^4\right)}-\sum _{k=2}^{\infty } \frac{2 e^{-i \pi k} x^4 \left(x^2+\pi \right)}{\pi ^2 \left(k^2-1\right) \left(\pi ^2 k^2-x^4\right)}-\frac{x^2}{2 \pi ^2}+\frac{1}{x^2}+\frac{2 x^2}{\pi ^2-x^4}=\csc \left(x^2\right)$$
$$-\sum _{k=1}^{\infty } \frac{\pi (-1)^{-k}}{(\pi -k) \left(\pi k-x^2\right)}+\sum _{k=1}^{\infty } \frac{(-1)^{-k} x^2}{\pi (\pi -k) \left(\pi k-x^2\right)}+\frac{1}{x^2}-\frac{1}{\pi ^2}+\csc \left(\pi ^2\right)=\csc \left(x^2\right)$$
I have calculated a new expansion around -Pi, Pi (thanks to the comments of @Jean Marie)and the almost perfect approximation can be seen graphically using the double poles of the function it seems that Mittag leffler's theorem can be extended to functions with double poles,although I have not been able to prove equality itself
of course you can always calculate the expansion of two points around -Pi,Pi of Csc(x) and change x by x^2, and this if equality can be proved, (well I don't know if these developments have been done before , but they do not seem to have much application in mathematics due to the interest they arouse) $$\sum _{k=2}^{\infty } \frac{(-1)^{-k} \left(\pi -x^2\right) \left(x^2+\pi \right)}{\pi ^2 (k-1) (k+1) \left(\pi k-x^2\right)}-\sum _{k=2}^{\infty } \frac{(-1)^{-k} \left(\pi -x^2\right) \left(x^2+\pi \right)}{\pi ^2 (k-1) (k+1) \left(\pi k+x^2\right)}+\frac{\left(3 \pi -x^2\right) \left(x^2+\pi \right)}{4 \pi ^2 \left(\pi -x^2\right)}-\frac{\left(x^2-\pi \right) \left(x^2+\pi \right)}{\pi ^2 x^2}-\frac{\left(\pi -x^2\right) \left(x^2+3 \pi \right)}{4 \pi ^2 \left(x^2+\pi \right)}=\csc \left(x^2\right)$$