Find the point on the curve farthest from the line $x-y=0$. the curve $x^3-y^3=1$ is asymptote to the line $x-y=0$. Find the point on the curve farthest from the line $x-y=0$.can someone please explain it to me what the question is demanding?
I cant think it geometrically as I am not able to plot it
Also is there any software to plot such graphs.?

Rotate by $-\pi/4$. This corresponds to $x\mapsto x-y$ and $y\mapsto x+y$ (up to a scalar factor). Then your line becomes $x-y-(x+y)=0\iff y=0$ and the equation of the curve becomes
$$(x-y)^3-(x+y)^3=1\iff -2y^3-6yx^2=1\iff x^2=-\frac{1+2y^3}{6y}$$
as $y$ can't be $0$. You want to find the farthest point from the line $y=0$, so you need to minimize $y$ so that the fraction on the right is positive. So the solution is $(0,-1/\sqrt[3]2)$, which is
$$(1/\sqrt[3]2,-1/\sqrt[3]2)$$ in the original coordinates.
