= Solution
Put $A=\wp(w)$ and $B=\wp'(w)$. At $z=0$, the first column has $\wp(z)=z^{-2}+O(z^2)$ and $\wp'(z)=-2z^{-3}+O(z)$, whereas the third has $\wp(-z-w)=A+Bz+O(z^2)$. Expanding the <determinant>, the coefficient multiplying $\wp'(z)$ is $\wp(-z-w)-A=O(z)$, so the apparent cubic pole reduces to order at most two. The only other possible pole is at $z=-w$, and $f(-z-w)=-f(z)$ gives the same bound there. Thus, \b[if $f$ is nonconstant, its total pole order and valency are at most four]. A constant case is handled by finding a zero below.
Column coincidences give zeros at $z=w$, at $z=-2w$, and at the four solutions of $2z=-w$ on $\mathbb C/\Lambda$. The latter are distinct because $\Lambda/2\Lambda$ has four elements. If $3w\notin\Lambda$, the first two points are distinct from each other and from these four solutions.
One exceptional pole issue needs care: when $2w\in\Lambda$, the first two prospective zeros are $w$ and $0$, the possible pole locations. They are actually removable zeros. Here $B=0$, and, with $C=\wp''(w)$, evenness about this half-period gives
$$
\wp(-z-w)=A+\frac C2z^2+O(z^4),\qquad
\wp'(-z-w)=-Cz+O(z^3).
$$
The <determinant> becomes $(A-\wp(z))\wp'(-z-w)+\wp'(z)(\wp(-z-w)-A)$. Its two $C/z$ contributions cancel and it is $O(z)$; reflection transfers this conclusion to $z=w$. Thus there really are \b[six distinct zeros] also in this case.
If $3w\in\Lambda$ but $w\notin\Lambda$, then $z=w=-2w$ is one of the four solutions of $2z=-w$. All three columns coincide there, with no pole. Both the <determinant> and its first derivative vanish, since each term in the derivative still has two identical columns. Hence there are \b[four distinct zeros, at least one multiple], giving at least five zeros counted with multiplicity.
For a nonconstant <elliptic function>, zeros and poles in a period parallelogram have equal total multiplicity, by the <argument principle>. Either count contradicts total pole order at most four. A constant with a zero is zero, so
$$
\boxed{f\equiv0.}
$$
This is the <elliptic collinearity determinant>. Consequently, when $z_1+z_2+z_3\in\Lambda$, the three vectors $(1,\wp(z_i),\wp'(z_i))$ have zero <determinant>. The three points on the <Weierstrass equation of an elliptic curve> are therefore \b[collinear in $\mathbb C^2$].
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