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.?

What the question is demanding is this: the curve gets closer and closer to the line; so it makes sense that there’s a point on the curve that’s farthest from the line.
I’ll show you two ways of doing this: the first is special, the second is general.
The first is to notice, once you’ve drawn the picture, that both $x^3-y^3=1$ and $x-y=0$ are symmetric about the line $y=-x$. the explicit description of the reflection is $(x,y)\mapsto(-y,-x)$. It makes sense that the one point on the curve that’s fixed with respect to this symmetry will have all sorts of special properties. Maybe a local minimum of distance to the line, but much more likely a maximum of this distance. With your picture, you see that it must be a maximum. What is this point? It’s the point on the curve that’s on the line of symmetry $y=-x$. So you make that substitution, get $2x^3=1$, $x=1/\root3\of2$, so the point is $(1/\root3\of2,-1/\root3\of2)$.
The other, general, method, is to use the fact that the closest and farthest points to a straight line have the property that their tangent is parallel to the line (assuming your curve is differentiable, of course). This is a generalization of the fact that the points on a graph $y=f(x)$ that are closest and farthest from the $x$-axis have tangents that are parallel to the $x$-axis (i.e. $f'(x)=0$ there). So you just want a point on your curve where the tangent has slope $1$, same as your line.
That’s easy: differentiating implicitly, you get $3x^2dx-3y^2dy=0$, $dy/dx=x^2/y^2$, and for this quantity to be equal to $1$, you need either $x=y$ or $x=-y$, The former is impossible on your curve, so $x=-y$, and you get the same values for the coordinates as above.
You can do it with Lagrange Multipliers, too, but it boils down to the “general” method I just showed you.
