= Solution
An <isogeny of elliptic curves> is a nonconstant morphism preserving identity points; it is automatically a finite surjective group homomorphism. On the affine chart $v\ne0$, put
$$
t=\frac uv,
\qquad s=\frac wv.
$$
The equation of $E$ is $t^3+d=s^3$, and the proposed map is
$$
x=ts=\frac{uw}{v^2},
\qquad y=t^3=\frac{u^3}{v^3}.
$$
It lands on $E'$ because
$$
y^2+dy=t^6+dt^3=t^3(t^3+d)=t^3s^3=x^3.
$$
The rational formulas extend across $v=0$ to a morphism sending $O_E$ to $O_{E'}$. It is nonconstant, hence an isogeny. On function fields, $t$ satisfies $t^3=y$, so the degree is at most three; generically the three cube roots give three distinct preimages. Equivalently, the points with $v=0$ form its three-element geometric kernel. Therefore
$$
\boxed{\deg\phi=3.}
$$
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