Here's the trick:

$u(x)$ will evaluate to 1 if $x$ is positive and zero otherwise, whatever be the expression you have in the place of x.

So, for $u(4-t)$, we need to see what values of $t$ keep $(4-t)$ positive...

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.

My book on ideals and varieties calls this the 'twisted cubic'.

More precisely I'm talking about the book Ideals, Varieties, and Algorithms where on page 20 it is mentioned that in the context of the variety $\textbf{V}(y-x^2,z-x^3)$ (which has previously been introduced as being the twisted cubic):

"... Note that setting $x=t$ in $y-x^2=z-x^3=0$ gives us a parametrization $$x=t$$ $$y=t^2$$ $$z=t^3$$ of the twisted cubic."