Is it simul?

$\large x - y = 3 \\ \large x^2 - y^2 = 63\\ \Rightarrow (x+y)(x-y)=63\\ \therefore x+y=21$

Afterthat, $x^2+y^2=\frac{(x+y)^2+(x-y)^2}{2}=\frac{21^2+3^2}{2}=\frac{450}{2}=\boxed{225}$

$\begin{aligned} x^2 - y^2 &= 63 \\ \implies (x+y)(x-y) &= 63 \\ \implies (x+y)(3) &= 63 \\ \implies (x+y) & = 21 \end{aligned}$

\[\begin{cases} x - y = 3 \\ x + y = 21

\end{cases}

\implies x = 12, y = 9\]

$x^2 + y^2 = 81 + 144 = \boxed{225}$