Limit of (x^3-y^3)/(x^2+y^2) at the Origin

The axes and the diagonal

Start with the two axes and the diagonal $y = x$, since each of the three turns $f$ into a function of $t$ alone.

The diagonal gives exactly zero at every $t$, not just in the limit, which looks like unusually strong evidence. It is not. Strength of agreement along one curve says nothing about the curves you did not try, and here the numerator was cancelled by the symmetry of the path before the limit was ever taken.

Polar coordinates and the triangle inequality

Substituting $x = r\cos\theta$ and $y = r\sin\theta$, the numerator is degree three in $r$ and the denominator is degree two, so one factor of $r$ survives.

Bound the bracket with the triangle inequality: $\left|\cos^3\theta - \sin^3\theta\right| \le \left|\cos^3\theta\right| + \left|\sin^3\theta\right| \le 1 + 1 = 2$.

Both outer expressions tend to $0$ as $r \to 0$, so $f$ is squeezed to $0$. The constant $2$ is generous; the sharp value of $\max\left|\cos^3\theta - \sin^3\theta\right|$ is $1$, attained at $\theta = 0$ and $\theta = \frac{\pi}{2}$. You do not need the sharp constant, because $Cr \to 0$ for any fixed $C$.

The mistake: running the triangle inequality backwards

A frequent slip is to bound the numerator by writing $\left|x^3 - y^3\right| \le \left|x^3\right| - \left|y^3\right|$, or to drop the second term entirely on the grounds that it is small. The triangle inequality points the other way: $|a - b| \le |a| + |b|$. The reversed version is false, and it can hand you a quantity that does not bound anything.

The second half of the mistake is finding a bound that still depends on $\theta$ and calling it done. A bound is only useful for a squeeze when the thing you compare against tends to zero on its own, with no reference to direction. Here that thing is $2r$, and $r$ is the distance to the origin.