= Solution
The inequalities imply
$$
B_{d_1}(x,r/\beta)\subseteq B_{d_2}(x,r)
\quad\text{and}\quad
B_{d_2}(x,\alpha r)\subseteq B_{d_1}(x,r).
$$
Thus the two metrics induce the same <open sets> and are equivalent.
The converse fails. On $\mathbb R$, let
$$
d_1(x,y)=|x-y|,
\qquad
d_2(x,y)=|\arctan x-\arctan y|.
$$
The metrics are equivalent because $\arctan:\mathbb R\to(-\pi/2,\pi/2)$ is a homeomorphism, but
$$
\frac{d_2(0,n)}{d_1(0,n)}\longrightarrow0,
$$
so no positive lower comparison constant exists. This is an <equivalent metrics need not be bi-Lipschitz equivalent> example.
Equivalent metrics on the codomain also need not give the same <uniform convergence>. Take $X=\mathbb N$, $Y=\mathbb R$, and use the two metrics above. Define
$$
f(k)=k,
\qquad
f_n(k)=\begin{cases}2n,&k=n,\\k,&k\ne n.
\end{cases}
$$
Then
$$
\sup_kd_2(f_n(k),f(k))
=\arctan(2n)-\arctan n\longrightarrow0,
$$
so $f_n\to f$ uniformly for $d_2$, whereas
$$
\sup_kd_1(f_n(k),f(k))=n,
$$
so convergence is not uniform for $d_1$. This is the <equivalent codomain metrics need not preserve uniform convergence> phenomenon.
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