Let $f$ be a function from a subset of $\mathbb{R}^2$ to $\mathbb{R}^3$. The graph of $f$ generally defines a surface. Can we find a function $F : A \subset \mathbb{R}^3 \to \mathbb{R}$ such that the parametrized surface defined by $f$ is exactly the surface $F(x,y,z)=0$ ?
One can assume regularity hypotheses if necessary.
Useful keyword: elimination theory.
Let's look at this one dimension down. Then you have $x=X(t)$ and $y=Y(t)$ as your parametrization. You can also write this as $X(t)-x=0$ and $Y(t)-y=0$. Now if $X$ and $Y$ are polynomials in $t$, then you can combine both equations and eliminate $t$ in the process. One way of doing so would be computing the resultant of the two left hand sides. The resultant is zero if both terms have a common zero, i.e. if there exists a $t$ satisfying both equations. Thus using the resultant of $X(t)-x$ and $Y(t)-y$ with respect to the variable $t$ will give you a suitable $F$.
In higher dimension you can do something similar. You have three equations in two variables, i.e. $X(t,u)-x=Y(t,u)-y=Z(t,u)-z=0$. You can combine pairs of these to eliminate one variable, obtaining two equations in one variable just as above. Actually you'd get a third equation from the third possible pair, but that will be adding no information. Then combine the resulting equations to eliminate the second variable. There is a chance of this process adding spurious solutions, where the two equations from the first step can only be satisfied by two different choices for the first variable. So you might want to factor the resulting implicit description to see whether it contains components not covered by the parametric form. Or you could use Gröbner bases instead of resultants, but I have less experience doing that.
If $X,Y,Z$ are not polynomials but rational functions (i.e. polynomial divided by polynomial), you might get a suitable implicit form by bringing the whole left hand side to a common denominator and then considering the full numerator. If your functions are not even rational, things will become more complicated as you likely left the well understood scope of algebraic surfaces.