 Solution

Take $\varphi =\forall x_1(x_0=x_1)$. The term $x_2$ is free for substitution for $x_0$: its variable is not bound by the enclosing quantifier. The term $x_1$ is not free for substitution for $x_0$, because insertion under $\forall x_1$ captures that occurrence.

Thus the first substitution gives $\forall x_1(x_2=x_1)$; the second, illegitimate substitution would give $\forall x_1(x_1=x_1)$.