Sketch a curve given parametrically by $x=2t-4t^3$ and $t^2-3t^4$ I am unable to see how to eliminate $t$. Wolfram Alpha fails at it too. 
$$x=2t-4t^3$$ $$y=t^2-3t^4$$
I can guess that the curve is a polynomial equation so in principle I can write this as 
$$w_1 x^4 +w_2 x^3 +w_3 x^2 +w_4 x +w_5 y^3 +w_6 y^2 +w_7 y +w_8=0$$
for the powers of $t$ to cancel when plugged in I am determine all these $w's$ by substituting the $t's$ for $x(t)$ and $y(t)$ and setting each of the new coefficients of the $t$'s $C_i(w_1,...w_8)$ to zero individually. I have my curve determined if I put each of the $C_i(w_1,...w_8)=0$ and solve for the $w$'s. 
All this was in principle, but this is like an examination question so there must be a clever manipulation/way, which I have been unable to find.
 A: My opinion is that you should not try to eliminate $t$ at all. Instead, you should think about how $x$ and $y$ behave as $t$ varies, find some important points such as the critical points with respect to $x$ and $y$, and try to sketch the graph from that information.
$x=2t-4t^3$ means that as $t$ goes from $-\infty$ to $\infty$, $x$ decreases from $\infty$ to a local minimum of $-\sqrt{8/27}$ at $t = -1/\sqrt{6}$, then rises to $\sqrt{8/27}$ at $t = 1/\sqrt{6}$, then falls again to $-\infty$.
$y=t^2-3t^4$ means that $y$ starts and ends at $-\infty$, with local maxima of $1/12$ when $t = \pm 1/\sqrt{6}$, and a local minimum of $0$ at $t = 0$.
Interestingly, $t = \pm 1/\sqrt{6}$ is a critical point for both $x$ and $y$, both of whose derivatives change sign, so the curve forms a cusp at that point.
That's enough information to follow the curve as $t$ goes from $-\infty$ to $\infty$: it comes in from the bottom right quadrant $(\infty,-\infty)$, goes upward and leftward till it hits the cusp $(-\sqrt{8/27},1/12)$, upon which it turns around, passing smoothly through $(0,0)$ where its tangent is horizontal, until it reaches the second cusp $(\sqrt{8/27},1/12)$, and then turns around a second time and exits at the bottom left towards $(-\infty,-\infty)$.
This is consistent with the plot arising from Juan Joder's answer.
A: I wouldn't know how to do it in an exam, but using a Gröbner basis routine such as that in Mathematica easily gives the implicit Cartesian equation
$$27x^4-4x^2(36y+1)+16y(4y+1)^2=0$$
A: Natural points for sketching include: when $x=0$, giving $t = 0 \text{ or } \pm 1/\sqrt{2}$ and so $y = 0 \text{ or } -1/4$; and when $y=0$, giving $t = 0 \text{ or } \pm 1/\sqrt{3}$ and so $x = 0 \text{ or } \pm 2/\sqrt{27}$.  So these four points suggest something interesting is going on near the origin.     
Just drawing the curve parametrically using Excel gives something like 
but zooming in gives the more interesting 
largely matching Rahul Narain's answer and Wolfram Alpha
A: You may have a look at Gibson's book "Elementary Geometry of Algebraic Curves",
in particular discussions on page 9 and page 178.
As to your problem at hand, you just consider x and y as parameters in equations
$f(t) =-4t^3 +2t -x =0$
and 
$g(t) =-3t^4 +t^2 -y =0$.
When x and y are chosen properly there is a value of t solving both equations.
That means there is a common root for both polyomials. So to find the relationship
between x and y you need only to solve the "resultant" of these two polynomials
f(t) and g(t). This resultant is a determinat involving x and y which vanishes
if and only if there is such a common root. Hence the resultant $R(f,g)=0$ gives the
desired equation relating x and y.
As to your original method, you overlooked the possibility that the equation relating x and y might have monomials like $xy$ and $x^2y$. Thus it is destined to fail.
If my computation is correct, the final equation should be
$$\frac{27}{64}x^4-\frac{1}{16}x^2-\frac{9}{2}x^2y+4y(y+\frac{1}{4})^2=0.$$
