I want to prove that:
$$\wp(2z) = \frac14 \left(\frac{\wp''(z)}{\wp'(z)}\right)^2 - 2\wp(z)$$
and I am asked to do it using the poles of both sides. I think that $\wp(2z)$ has exactly 4 poles, at the half lattice points. and the right side kind of looks like it has 4 poles in the same places too, because those are the zeroes of $\wp'$, but $\wp$ has a pole at $0$, so the singularity at zero could be removable, not sure how to prove it isn't.
But even then, just because they have the same poles doesn't tell me much (I think?).
I would express the functions involved in power series.
$$\wp(2z) \!=\! \frac{1}{4z^2}\!+\!O(z)^2\!,\, 2\wp(z) \!=\! \frac{2}{z^2}\!+\!O(z)^2\!,\, \wp''(z) \!=\! \frac{6}{z^4}\!+\!O(z)^0,\, \wp'(z) \!=\! -\frac{2}{z^3}\!+\!O(z)^1\!, \tag{1} $$ and now $$\wp(2z) + 2\wp(z) =\frac{9}{4z^2}+O(z)^2,\quad \frac14 \left(\frac{\wp''(z)}{\wp'(z)}\right)^2 = \frac{9}{4z^2}+O(z)^2, \tag{2}$$ so your equation is true up to $\;O(z)^2\;$ for the pole at the origin in the period parallelogram. Similar considerations apply at the other three poles at half lattice points. Now the two sides have the same poles and residues which implies that the difference of the two sides has no poles, and since $\wp$ has a period parallelogram, by Liouville's theorem the two sides must differ by a constant. Using equation $(2)$ that constant is zero.