Extreme values in a closed, bounded regions will either be at a local extrema or on the boundary. So you need to check the values of the function along the boundary of the region. You do this by reducing it to a one variable equation on the boundary.
For instance, you have $x=0, y\in [0,1]$ is one boundary, so plugging in $x=0$ has you finding the extreme values of $-y^3$ in $[0,1]$. Again, the same rule applies, extreme values will occur at a local extrema or at an endpoint, so you'll check at $y=0,y=1$ and anywhere the derivative is 0.
Your other boundaries are $y=0, x\in [0,1]$, and $x+y=1$. We can solve for either x or y to reduce it to a one variable equation. For instance, you could plug in $y=1-x$, reduce it to an equation in $x$ with again boundary values of $[0,1]$
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Check the corners if you are finding global extrema in a closed domain. The four corners of the rectangular boundary must also be considered, just as how the two endpoints of a domain in single-variable calculus must be considered. Every extrema inside the domain and on the boundary of the domain, with the addition of the four corners, must be plugged into the function to determine global extrema. Below, we list the locations of the global maximum and minimum. They have values of f≈±6.041,{\displaystyle f\approx \pm 6.041,}
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