One way we could go is to subtract $x$ times the second equation from the first, getting us: $$xy^2 - 2y + 3x^2-x(x^2y + 2x + y^2) = 0\\xy^2 - 2y + 3x^2-x^3y - 2x^2 -xy^2 = 0\\x^2-x^3y-2y=0\\x^2=x^3y+2y\\x^2=(x^3+2)y$$ Note that if we had $x^3+2=0,$ then the last equation above would become $x^2=0,$ so $x=0,$ contradicting our assumption that $x\ne 0.$ Thus, we don't have to worry about $x^3+2$ being zero, and so $$y=\frac{x^2}{x^3+2}.\tag{$\star$}$$ Substituting for $y$ in the second equation gets us $$x^2\cdot\frac{x^2}{x^3+2}+2x+\left(\frac{x^2}{x^3+2}\right)^2=0,$$ or $$x\left(\frac{x^3}{x^3+2}+\frac{x^3}{(x^3+2)^2}+2\right)=0,$$ which (since $x\ne 0$) is equivalent to $$\frac{x^3}{x^3+2}+\frac{x^3}{(x^3+2)^2}+2=0\\x^3(x^3+2)+x^3+2(x^3+2)^2=0\\x^6+3x^3+2(x^6+4x^3+4)=0\\3x^6+11x^3+8=0\\3\left(x^3\right)^2+11x^3+8=0\\x^3=\frac{-11\pm\sqrt{121-96}}6\\x^3=\frac{-11\pm \sqrt{25}}6\\x^3=\frac{-11\pm 5}6\\x^3=-1\textrm{ or }x^3=-\frac83$$
If we're looking for real solutions, then we have $x=-1$ or $x=-\frac{2}{\sqrt[3]3},$ but if we're interested in non-real solutions, too, then since $-\frac12\pm\frac{\sqrt3}2$ are cube roots of $-1,$ too, we also have $x=\frac12\pm\frac{\sqrt3}2$ and $x=-\frac{2}{\sqrt[3]3}\left(-\frac12\pm\frac{\sqrt3}2\right).$ (For brevity, I will denote $\omega=-\frac12+\frac{\sqrt3}2$ and $\overline\omega=-\frac12-\frac{\sqrt3}2.$)
When $x\in\left\{-1,\omega,\overline\omega\right\},$ we have $x^3=-1,$ so $x^3+2=1,$ and so $(\star)$ becomes $$y=x^2.$$ Thus, we obtain the solutions $(-1,1),$ $\left(\omega,\omega^2\right)=\left(\omega,-\overline\omega\right),$ and $\left(\overline\omega,\overline\omega^2\right)=\left(\overline\omega,-\omega\right).$
When $x\in\left\{-\frac{2}{\sqrt[3]3},-\frac{2\omega}{\sqrt[3]3},-\frac{2\overline\omega}{\sqrt[3]3}\right\},$ we have $x^3=-\frac83,$ so $x^3+2=-\frac23,$ and so $(\star)$ becomes $$y=-\frac32x^2.$$ Thus, we obtain the solutions $\left(-\frac{2}{\sqrt[3]3},-2\sqrt[3]3\right),$ $\left(-\frac{2\omega}{\sqrt[3]3},2\overline\omega\sqrt[3]3\right),$ and $\left(-\frac{2\overline\omega}{\sqrt[3]3},2\omega\sqrt[3]3\right).$
In summary, our solutions are: $$(0,0),(-1,1),\left(-\frac{2}{\sqrt[3]3},-2\sqrt[3]3\right),\left(\omega,-\overline\omega\right),\left(\overline\omega,-\omega\right),\left(-\frac{2\omega}{\sqrt[3]3},2\overline\omega\sqrt[3]3\right),\left(-\frac{2\overline\omega}{\sqrt[3]3},2\omega\sqrt[3]3\right),$$ with the first three being the real solutions.
How does the imaginary number prevents you from differentiating, though the complex number always cancels out
Maybe you should check the trigonometric cubic formula
Speaking of your topic that's interesting, why should complex number appear when we don't need them?
Looking at the formula you posted $x = p + \sqrt[3]{q+Q} + \sqrt[3]{q-Q}$, the cube root part are conjugate and we know in normal cases, conjugates can be simplified
$(q+Qi)(q-Qi) = q^2+Q^2$
$(q+Qi)+(q-Qi) = 2q$
$(q+Qi)^2+(q-Qi)^2 = 2q^2-2Q^2$
$(q+Qi)^3+(q-Qi)^3 = 2q^3-6qQ^2$
$\sqrt{q+Qi}+\sqrt{q-Qi} = \sqrt{2q±2\sqrt{q^2+Q^2}}$
$\sqrt[3]{q+Qi}+\sqrt[3]{q-Qi} = ?$
the cubic formula can't be simplified, because it's already in its simplest case
check this cubic equation $x^3-6x-6=0$, $x = \sqrt[3]{2}+\sqrt[3]{4}$
Just like the quadratic formula makes use of the square root of unity, $\sqrt{1} = \pm 1$
So the truth behind this problem is that the cubic formula makes use of the cube root of unity $\sqrt[3]{1} = 1$, $-\frac{1}{2}+\frac{\sqrt{-3}}{2}$ and $-\frac{1}{2}-\frac{\sqrt{-3}}{2}$ meaning that it would always have complex number
In general every polynomial of degree $n$ makes use of the nth-root of unity in its formula
Best Answer
If you do have Mathematica, use the command
GroebnerBasis[{pol1, pol2}, {x,y},{a}]
where poli are the two expressions that equal $0$. This will eliminate the variable $a$.
On WolframAlpha I got a polynomial in $x$, $y$ with fairly large coefficients. Check this link here.
Hope there were no mistakes in the input.
$\bf{Added:}$. You will get the resultant of the two expressions considered as polynomials in $a$. The Groebner command is just more general, can eliminate some variables ( the ones in the second group) from a system of equations.