The curve $x^3-y^3=1$ is asymptotic t the line $x=y$.
Find the point in the curve farthest from the line $x=y$
This is just need of further details in this question.
Kindly give details about, if we are talking about asymtotic curve to the line $x=y$, what does it really means. And why we need the family of curves $y=x+C$ to look this problem(as in the answer given by Calvin).
On
$(x-y)\left(\frac34(x+y)^2+\frac14(x-y)^2\right)=1$
so $x-y$ is greatest when $(x+y)^2$ is smallest.
On
The distance from a point $(x_0,y_0)$ satisfying $x^3 - y^3 = 1$ to the line $y = x$ must be measured along a line segment perpendicular to $y = x$; that is to say, the foot of the perpendicular $(x_1, y_1)$ must satisfy the system $$x_1 = y_1, \quad y_1 - y_0 = -(x_1 - x_0),$$ or quite simply $$x_1 = y_1 = \frac{x_0 + y_0}{2}.$$ Therefore, the distance is given by $$D(x_0,y_0) = \frac{|x_0 - y_0|}{\sqrt{2}} = \frac{|(y_0^3+1)^{1/3} - y_0|}{\sqrt{2}}.$$ Taking the derivative of the numerator with respect to $y_0$ and locating the critical points gives $y_0 = -2^{-1/3}$, hence by symmetry $(x_0, y_0) = (\pm 2^{-1/3}, \mp 2^{-1/3})$ are the desired solutions, with maximal distance $2^{1/6}$.
\begin{align}& {\rm f}\left(\,x\,\right)\equiv x - \left(\,x^{3} - 1\,\right)^{1/3}\quad \mbox{always decreases becauses}\quad {\rm f'}\left(\,x\,\right)<0\quad\mbox{when}\quad x > 1. \\[5mm]& \left(\,1,0\,\right)\quad\mbox{is the farthest point from the line}\quad y = x. \end{align}