= Solution
Because $f$ is a bijection and $e$ is a metric,
$$
d'(x,y)=e(f(x),f(y))
$$
is nonnegative, symmetric, obeys the triangle inequality, and vanishes exactly when $x=y$. Hence it is a metric. Its open sets are precisely the inverse images under $f$ of $e$-open sets in $N$. Since $f:(M,d)\to(N,e)$ is a <homeomorphism>, these are exactly the $d$-open sets. Thus $d'$ and $d$ are <equivalent metrics>.
Let
$$
h(x)=\frac12+\frac1\pi\arctan x,
$$
a homeomorphism $\mathbb R\to(0,1)$, and define
$$
\boxed{d'(x,y)=|h(x)-h(y)|}.
$$
This metric induces the standard topology on $\mathbb R$. However, the sequence $x_n=n$ is $d'$-Cauchy because $h(n)\to1$. If it converged to $x\in\mathbb R$, then continuity of $h$ would give $h(x)=1$, impossible. Therefore $(\mathbb R,d')$ is not complete.
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