For $i=1,2,3$ let $P_i$ be real polynomials such that $P_3$ is positive. Define the mapping \begin{align*} f(x,y) = \left(x+\frac{P_1(x,y)}{1+\sqrt{P_3(x,y)}},y+\frac{P_2(x,y)}{1+\sqrt{P_3(x,y)}}\right), x,y \in \mathbb{R}. \end{align*}
Does $f$ map algebraic curves to algebraic curves?
Note: Numerically, I investigated many non-trivial examples and they all looked algebraic again.
Explaining my comment:
Assume that the original algebraic curve is given in parametric form $$x(t)=r_1(t), \ \ \ y(t)=r_2(t),$$ where $r_i$ are rational functions. Applying $f$ to it gives $$f(x(t),y(t))=(u,v)=\left(r_1(t) + \frac{r_3(t)}{1+\sqrt{r_5(t)}},r_2(t) + \frac{r_4(t)}{1+\sqrt{r_5(t)}}\right).$$
This can be rewritten as a polynomial equation system in $t$. Computation of the determinant $\det$ of its Sylvester matrix gives a polynomial equation $\det(u,v)=0$, such that $(u,v)$ is again an algebraic curve.
Problem here: As far as I know, an algebraic curve is rationally parameterizable if and only if its genus is equal to zero.