I am asked to show whether the relation
$$ f(x,y) = x^3 + y^3 - 3xy = 0, -\infty < x < \infty $$
is a solution to the equation
$$ F(x,y,y') = (y^2 - x)y' - y + x^2 = 0, -\infty < x < \infty $$
The difficulty I am having is not in implicitly differentiating $f(x,y)$, it's in seeing how it can be a solution on the interval. The relation does not define a function however, consider the following image :
The above shows three possible functions we can use for $ f(x,y) $ lets assume we are using the left most function. Implicitly differentiating $f(x,y)$ yields the following:
$$ 3x^2 + 3y^2y' - 3y - 3xy' $$ $$ = (y^2 - x)y' + x^2 - y = 0, x \in (-\infty < 2^{ \frac{2}{3} }) \cup (2^{\frac{2}{3} } < \infty) $$
Notice the domains don't match up, this is a subset of $(-\infty, \infty)$. Since the domains don't match up doesn't if follow $f(x,y)$ can't be a solution?
$$ f(x,y)=x^3+y^3 - 3xy=0\\ 3x^2 + 3 y^2 y' - 3xy' - 3y =0\\ (3y^2-3x)y'=3y-3x^2\\ y' = \frac{y-x^2}{y^2-x} $$
So for all points $(x,y)$ on $f(x,y)=0$, we have $y'$.
$$ F=(y^2-x)y'-y+x^2\\ $$
Assume we are on $f(x,y)=0$, then we can substitute the expression for $y'$.
$$ F=(y^2-x)\frac{y-x^2}{y^2-x}-y+x^2 = 0 $$
so all together if we assume we are on $f(x,y)=0$, we can conclude $F=0$.
More generally we might have been given:
$$ F=(y^2-x)y'-y+x^2 + (x^3+y^3-3xy)*g(x) $$
In that case assuming we were on $f(x,y)=0$ we would also get $F=0$, but we would have to use both the expression for $y'$ as well as the fact $f(x,y)=0$ again to get rid of the last term.