As the other answer says, your partial derivatives are $\frac{\partial z}{\partial x} = y, \quad \frac{\partial z}{\partial y} = x \, . $
What does this mean in terms of actual gameplay? It means that, if you currently have stats $x=x_0$, $y=y_0$, you should (approximately) value small changes in $x$ by multiplying them by $y_0$, and value small changes in $y$ by multiplying them by $x_0$.
For example, say that at some point in the game your stats are $x=0.5, y=0.25$ (you have a $50$% chance of scoring a critical, and it does one-quarter again as much damage as an ordinary hit). Then, if you're given the option of increasing $x$ by $0.04$ at the cost of decreasing $y$ by $0.01$, you should value this at $0.04(0.25)-0.01(0.5)=0.01-0.005=0.005$. Since this is positive, that option is beneficial.
Since your initial formula for $z$ was fairly simple, this is a little bit superfluous. You could just plug everything into the original $z$-formula. Your initial damage output would be $z_0=1+(0.5)(0.25)=1.125 \, .$ Your new damage output would be $z_1=1+(0.5+0.04)(0.25-0.01)=1.1296 \, .$ Again, since $z_1>z_0$, we can see that taking this new option is beneficial.
Note that $z_1-z_0=0.0046$: very close to, but not actually equal to, the $0.005$ we computed by using partial derivatives.
Why would you ever use partial derivatives at all, if you could just use the original function to get an exact answer? In this case, I would say you probably shouldn't unless you're in a big hurry. If $z$ were more complicated, though, it might save you a bunch of time, especially if you've got a bunch of different possible changes to try from the same starting point. Since all you care about is which change leads to the biggest (positive) shift in $z$, you won't need to worry that the your partial-derivative computations are approximations unless you have two different potential options that are really close in value.
One thing to be careful about: since $x$ is a probability, presumably it's capped at $1$ (or maybe at some lower value if D3 doesn't like the idea that all your hits could be criticals). So your formula for $z$ probably isn't universal. But it's pretty obvious what you should do when it breaks down (maximize $y$ at all costs, stop worrying about $x$).