83 Given the equation $x^{2}+2 x y+y^{2}+3 x+y=0$ represents a parabola on the $x O y$ plane, try to find the equation of its axis of symmetry.
Let $\xi=\frac{x+y}{\sqrt{2}}, \eta=\frac{-x+y}{\sqrt{2}}$, then
$$
x=\frac{\xi-\eta}{\sqrt{2}} \quad y=\frac{\xi+\eta}{\sqrt{2}}
$$

The original equation becomes
$$
2 \xi^{2}+2 \sqrt{2} \xi-\sqrt{2} \eta=0
$$

That is

Let
$$
\begin{array}{c}
\left(\xi+\frac{\sqrt{2}}{2}\right)^{2}-\frac{\sqrt{2}}{2} \eta-\frac{1}{2}=0 \\
X=\xi+\frac{\sqrt{2}}{2}=\frac{x+y+1}{\sqrt{2}} \\
Y=\eta+\frac{\sqrt{2}}{2}=\frac{-x+y+1}{\sqrt{2}}
\end{array}
$$

The equation becomes
$$
Y=\sqrt{2} X^{2} .
$$

The axis of symmetry is $X=0$ i.e.,
$$
x+y+1=0
$$