1. Given non-zero real numbers $x, y$ satisfy $(3 x+y)^{5}+x^{5}+4 x+y=0$, then $\frac{x}{y}=$
1. $-\frac{1}{4}$ Analysis: The conditional equation is transformed into $(3 x+y)^{5}+(3 x+y)=-\left(x^{5}+x\right)$, let $f(t)=t^{5}+t$, it is clear that $f(t)$ is a monotonic odd function, and $f(3 x+y)=-f(x)=f(-x)$, thus $3 x+y=-x, \frac{x}{y}=-\frac{1}{4}$.