For those who have the book, this is from McKean and Moll's book on elliptic curves, exercise 2.14.7. This section deals with the Weierstrass elliptic function with periods $\omega_1$ and $\omega_2$, and this section operates under the assumption that $e_1+e_2+e_3=0$.
I was able to show in exercise 6 that $\wp(x+\omega_1/2)=e_1+(e_1-e_2)(e_1-e_3)[\wp(x)-e_1]^{-1}$, and that similar inequalities hold for the other two half-periods $(\omega_1+\omega_2)/2$ and $\omega_2/2$. Now if we add the three results together, along with $\wp(x)$, we get $4\wp(2x)$ from exercise 5, and this is fine. Exercise 7 then asks ``for the corresponding product'', which I can only assume is $$\wp(x)\wp\left(x+\frac{\omega_1}{2}\right)\wp\left(x+\frac{\omega_1+\omega_2}{2}\right)\wp\left(x+\frac{\omega_2}{2}\right).$$ Note that this product has a pole of order two at $x=0$, and this pole only appears in the first factor of this product. However the exercise is claiming that the answer is $$[\wp'(x)]^2\cdot 16(e_1-e_2)^2(e_2-e_3)^2(e_3-e_1)^2,$$ which has a pole of order six at $x=0$. I take it that this pole order discrepancy implies that the two elliptic functions cannot be equal (or perhaps I have a fundamental misunderstanding of complex analysis?), implying that the book has yet another typo. Am I mistaken in my thought process? Or have people resolved this issue before?