Example 4 Find the largest constant $c$ such that for all real numbers $x, y$ satisfying $x>0, y>0, x^{2}+y^{2}=1$, we always have $x^{6}+y^{6} \geqslant c x y$.


When $x=y=\frac{\sqrt{2}}{2}$, the inequality also holds, then $c \leqslant \frac{x^{6}+y^{6}}{x y}=\frac{1}{2}$.
To prove that under the conditions $x>0, y>0, x^{2}+y^{2}=1$, the inequality $x^{6}+y^{6} \geqslant \frac{1}{2} x y$ always holds. (*) is equivalent to
$\left(x^{2}+y^{2}\right)\left[\left(x^{2}+y^{2}\right)^{2}-3 x^{2} y^{2}\right] \geqslant \frac{1}{2} x y$, which simplifies to $\quad 6(x y)^{2}+x y-2 \leqslant 0$.
Let $\mu=x y$, since $0<x y \leqslant \frac{x^{2}+y^{2}}{2}=\frac{1}{2}$, we have $0<\mu \leqslant \frac{1}{2}$.
Define $f(\mu)=6 \mu^{2}+\mu-2,0<\mu \leqslant \frac{1}{2}$, then $f(\mu)$ is increasing on $\left(0, \frac{1}{2}\right]$, and $f\left(\frac{1}{2}\right)=0$.

Therefore, on $\left(0, \frac{1}{2}\right]$, we have $f(\mu) \leqslant 0$, hence $6(x y)^{2}+x y-2 \leqslant 0$.
In conclusion, the maximum value of $c$ is $\frac{1}{2}$.