Solution:
The gradient is equal to the derivative of y with respect to x. By implicit differentiation,
\left(x^3\cdot\frac{\differentialD y}{\differentialD x}+3x^2y\right)+\left(3xy^2\cdot\frac{\differentialD y}{\differentialD x}+y^3\right)=0
x^3\cdot\frac{\differentialD y}{\differentialD x}+3x^2y+3xy^2\cdot\frac{\differentialD y}{\differentialD x}+y^3=0
x^3\cdot\frac{\differentialD y}{\differentialD x}+3xy^2\cdot\frac{\differentialD y}{\differentialD x}=-\left(3x^2y+y^3\right)
\left(x^3+3xy^2\right)\frac{\differentialD y}{\differentialD x}=-\left(3x^2y+y^3\right)
\frac{\differentialD y}{\differentialD x}=-\frac{3x^2y+y^3}{x^3+3xy^2}
The derivative becomes undefined if the denominator is zero. Therefore,
x^3+3xy^2=0
x\left(x^2+3y^2\right)=0
Splitting the factors,
x=0 and x^2+3y^2=0
For the second factor,
x^2=-3y^2
The equation does not hold for any pairs of x and y on real numbers EXCEPT when x=0 and y=0. Therefore,
x=0,y=0
Checking if the values exist in the curve:
0^3\cdot0+0\cdot0^3\stackrel{?}{=}27
0\ne27
Therefore, the ordered pair is not on the curve. Hence, the equation is undefined at the values of x=0 and y=0 due to the fact that there is a discontinuity in the said ordered pair.