Implicit differentiation involving a sliding ladder

$4/3 \text{ ft/min}$ and $-1 \text{ ft/min}$ are the instantaneous rates of change when $x = 3$ and $y = 4$. That rate of change is constantly changing as you pass that instant, and will not stay the same for a whole minute. Thus your analysis is incorrect because it assumes constant rates of change for a whole minute.

Your finding that $\frac{dx}{dt}=\frac{4}{3}$ relies on the fact that $y=4$ and $x=3$. This is true at $t=0$, but not true for any later time. As $x$ and $y$ change, also $\frac{dx}{dt}$ changes.

The issue here is that when you substitute for $x, y, \frac{dy}{dt}$ in your differentiated expression, you are substituting those values, like you state, "at some [particular] time," say, $t_0$: If we use notation that reflects this, we have $$2 x(t) x'(t) + 2 y(t) y'(t) = 0,$$ and substituting at our particular time $t_0$, we have $$2 x(t_0) x'(t_0) + 2 y(t_0) y'(t_0) = 0.$$ Now, our given data is $x(t_0), y(t_0), y'(t_0)$ (actually, we know that $y'$ is constant, but this isn't relevant), and substituting and rearranging gives $$x'(t_0) = \frac{4}{3} \textrm{ ft/min} .$$ In particular, this tells us the speed $x'$ only at the particular time $t_0$; indeed, $x'$ varies with time.