419. Solve the system of equations:

$$
\left\{\begin{array}{l}
\frac{x^{2}}{y^{2}}+2 \sqrt{x^{2}+1}+y^{2}=3 \\
x+\frac{y}{\sqrt{x^{2}+1}+x}+y^{2}=0
\end{array}\right.
$$
$\triangle$ Let's eliminate the irrationality $\sqrt{x^{2}+1}$ from the system. For this, we express $\sqrt{x^{2}+1}$ from the first equation:

$$
\sqrt{x^{2}+1}=\frac{3 y^{2}-x^{2}-y^{4}}{2 y^{2}}
$$

Now we need to substitute this expression into the second equation, first moving the irrationality from the denominator to the numerator:

$$
x+y\left(\sqrt{x^{2}+1}-x\right)=0
$$

We get:

$x+y\left(\frac{3 y^{2}-x^{2}-y^{4}}{2 y^{2}}-x\right)+y^{2}=0$

$x+\frac{3 y^{2}-x^{2}-y^{4}-2 x y^{2}}{2 y}+y^{2}=0$

$2 x y+3 y^{2}-x^{2}-y^{4}-2 x y^{2}+2 y^{3}=0$,

$x^{2}+2 x\left(y^{2}-y\right)+y^{4}-2 y^{3}-3 y^{2}=0$.

We will consider the last equation as a quadratic equation in terms of $x$. Let's find its discriminant:

$$
\frac{1}{4} D=\left(y^{2}-y\right)^{2}-y^{4}+2 y^{3}+3 y^{2}=4 y^{2}
$$

Then

$$
x=y-y^{2} \pm 2 y ; \quad x_{1}=3 y-y^{2}, \quad x_{2}=-y^{2}-y
$$

1) Let $x=3 y-y^{2}$. Substitute this expression for $x$ into the formula for $\sqrt{x^{2}+1}$:

$$
\begin{aligned}
& \sqrt{\left(3 y-y^{2}\right)^{2}+1}=\frac{3 y^{2}-\left(3 y-y^{2}\right)^{2}-y^{4}}{2 y^{2}} \\
& \sqrt{\left(3 y-y^{2}\right)^{2}+1}=\frac{3-(3-y)^{2}-y^{2}}{2} \\
& \sqrt{\left(3 y-y^{2}\right)^{2}+1}=-y^{2}+3 y-3 \\
& 9 y^{2}-6 y^{3}+y^{4}+1=y^{4}+9 y^{2}+9-6 y^{3}+6 y^{2}-18 y \\
& 6 y^{2}-18 y+8=0,3 y^{2}-9 y+4=0
\end{aligned}
$$

From this, $y=\frac{9 \pm \sqrt{33}}{6}$. But both these values of $y$ do not satisfy the last irrational equation, as in this case its right-hand side is negative.

2) Let $x=-y^{2}-y$. Similarly, we find the pair $(0 ;-1)$. It satisfies the original system of equations.

Answer: $(0 ;-1)$.