A more general proof:
Let Q and R be the points at which lines through $P=(x_1,y_1)$ touch a non degenerate conic $S(x,y) \equiv Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$. In other words, lines PR and PQ are the tangents to this conic at points Q and R, and RQ is the chord of contact.
Let $PR(x,y)=0$, $PQ(x,y)=0$, $RQ(x,y)=0$ be the equation of these lines.
As RQ is polar of P in relation to this conic,
$$RQ(x,y)\equiv (Ax+By+D)x_1+(Bx+Cy+E)y_1+(Dx+Ey+F)=0$$
On the other hand, the equation $\lambda(PR(x,y).PQ(x,y))+\mu(RQ(x,y))^2=0$ represents all conics which are touched by lines PR and PQ at points R and Q. Therefore, for especific values of $\lambda$ and $\mu $ (none of which can be equal to zero, because otherwise S would be a degenerate conic):
$$S(x,y)\equiv \lambda(PR(x,y).PQ(x,y))+\mu(RQ(x,y))^2=0$$
Then,
$$S(x_1,y_1)=\lambda(PR(x_1,y_1).PQ(x_1,y_1))+\mu(RQ(x_1,y_1))^2,$$
$$S(x_1,y_1)=\mu(RQ(x_1,y_1))^2$$
Besides that,
$$RQ(x_1,y_1)=(Ax_1+By_1+D)x_1+(Bx_1+Cy_1+E)y_1+(Dx_1+Ey_1+F),$$
$$RQ(x_1,y_1)=S(x_1,y_1)$$
Thus
$$S(x_1,y_1)=\mu(S(x_1,y_1))^2,$$
$$\mu=\frac {1}{S(x_1,y_1)}$$
Therefore
$$S(x_1,y_1).S(x,y)\equiv S(x_1,y_1)\lambda(PR(x,y).PQ(x,y))+(RQ(x,y))^2,$$
$$S(x_1,y_1)\lambda(PR(x,y).PQ(x,y))\equiv S(x_1,y_1).S(x,y)-(RQ(x,y))^2$$
Finally, equating left and right members of this identity to zero, we get that the equation of tangents PR and PQ to conic S can be represented by equation
$$S(x_1,y_1).S(x,y)-(RQ(x,y))^2=0$$
Let the point of tangency be $(x_1,y_1)$
Then the equation of tangent is $\displaystyle\frac{y-y_1}{x-x_1}=3x_1^2-6x_1-7$.
When $y=0$, $\displaystyle x=x_1+\frac{-y_1}{3x_1^2-6x_1-7}$.
When $x=0$, $\displaystyle y=y_1-x_1(3x_1^2-6x_1-7)$.
\begin{align}
-2\left(x_1+\frac{-y_1}{3x_1^2-6x_1-7}\right)&=y_1-x_1(3x_1^2-6x_1-7)\\
-2\left(\frac{x_1(3x_1^2-6x_1-7)-y_1}{3x_1^2-6x_1-7}\right)&=y_1-x_1(3x_1^2-6x_1-7)\\
2&=3x_1^2-6x_1-7\\
3x_1^2-6x_1-9&=0\\
x_1&=-1\quad\text{or}\quad3
\end{align}
When $x_1=-1$, $y_1=(-1)^3-3(-1)^2-7(-1)+6=9$ and the $y$-intercept is
$$9-(-1)[3(-1)^2-6(-1)-7]=11>0$$
When $x_1=3$, $y_1=(3)^3-3(3)^2-7(3)+6=-15$ and the $y$-intercept is
$$-15-(3)[3(3)^2-6(3)-7]=-21<0$$
So, the point of tangency is $(-1,9)$.
Best Answer
At this point it's basically solving a system of two nonlinear equations in two variables. The equations are what you labeled $(1)$ and $(2)$.
From $(2)$ we can remove a factor of $2$ on the right and clear denominators to get: $$-y_1^2 + 4y_1 - 4 = 3x_1^2 - 3x_1^3$$ And we want to solve this simultaneously with: $$2x_1^3 = y_1^2 - 4y_1 + 8$$ In the first equation above, let's rewrite the LHS as $$-y_1^2 + 4y_1 - 4 = -(y_1^2 - 4y_1 + 8) + 4 = -2x_1^3 + 4.$$ Now solve that first equation above for $x_1$, using $-2x_1^3 + 4$ as the new LHS.
Can you take it from here?