Second system of equations I've solved that system and recieved $y=\frac{1}{4}$, $x = -\frac{4}{5}$ but there are one more pair of $(x, y)$ in book and I don't know what I have to do to find its.
\begin{cases} 2x - 3xy + 4y=0 \\ x + 3xy -3x = 1 \end{cases}
In book there are two answers and the second answer is $(1, -2)$.
I've reached $y=\frac{1}{4}$, $x = -\frac{4}{5}$ through that way:
\begin{array}{lcl}2x - 3xy + 4y = 0\\2x - 3xy = -4y\\x(2 - 3y) = -4y\end{array}
2)
\begin{array}{lcl}x + 3xy - 3x = 1\\-2x + 3xy = 1\\-x(2 - 3y) = 1\end{array}
Get new system:
\begin{cases} x(2 - 3y) = -4y \\ -x(2 - 3y) = 1 \end{cases}
Then I divide first to second and get
\begin{array}{lcl}-1 = -4y\\y = \frac{1}{4}\end{array}
Next solve one of equation and get \begin{array}{lcl} x = -\frac{4}{5}\end{array}
But how to get $x = 1, y = -2$?

In case you made a typo and the 2nd equation is actually $x+3xy-3y=1$, that equation can be factored:
$$
x-1 + 3y(x-1) = 0
$$
$$(x-1)(3y+1)=0$$
Which means that either $x=1$ or $y=-\frac{1}{3}$
Substitute that into the first equation to get the 2 pairs.
