Can someone clarify what it means to find the derivative "with respect to $t$"?
For example $$x^2 + y^2 = 625.$$
Using implicit differentiation I get: $2x + 2y(dy/dx) = 0$.
However with differentiation "respect to $t$" the answer is: $2x(dx/dt) + 2y(dy/dt) = 0$
Does it just mean multiplying it with $d/dt$ after differentiating it normally?
It means you have to think of $x$ implicitly as a function of $t$ rather than an independent variable. Same thing with $y$. So we can rewrite your original equation as $x(t)^2 + y(t)^2 = 625$, then just differentiate both sides of the equation with respect to $t$. The right hand side will become $0$ after differentiating. The derivative of $x(t)^2$ is, by the chain rule, $2x(t)\cdot (dx/dt)$ and similarly for $y(t)^2$, so the equation becomes $2x(t)\cdot (dx/dt) + 2y(t)\cdot (dy/dt) = 0$.