Let $A=\{(x,y) \in \mathbb{R}^2 : x^3 + y^2 = 2xy\}$ and let $B=\{(x,y) \in A : x>0, y>0 \}$. I want to parametrize $B$ as a curve $f(t)$. I know by the Implicit Function Theorem this can be done here.
Can anyone suggest a good place to start with this?
I'd start with a look at the plot. Seems as though you only have one function value per direction, so you could describe this in polar coordinates:
$$x=r\cos\varphi\qquad y=r\sin\varphi$$
I'm a big fan of rational functions and the tangent half-angle substitution, so make this
$$x=r\frac{1-t^2}{1+t^2}\qquad y=r\frac{2t}{1+t^2}$$
Plug these into your implicit equation. Find the common denominator, then look at the resulting numerator. It turns out to be
$$r^2\bigl((t^6 - 3t^4 + 3t^2 - 1)r - 4(t^5 + t^4 + t^2 - t)\bigr)=0$$
The first factor is spurious. The second tells you that you can easily solve for $r$:
$$r=\frac{4(t^5 + t^4 + t^2 - t)}{t^6 - 3t^4 + 3t^2 - 1} =4\frac{t(1+t^2)(1-t-t^2)}{(1-t^2)^3}$$
So from $t$ you can compute $r$ and from these compute $x$ and $y$:
$$x=4\frac{t(1-t-t^2)}{(1-t^2)^2}\qquad y=8\frac{t^2(1-t-t^2)}{(1-t^2)^3}$$
The portion of the curve which lies in the first quadrant starts and ends in the origin, i.e. with $r=0$. So compute the roots of the numerator of $r$ to find out the corresponding parameter values. You may then pick
$$0 to stay in the first quadrant. Done. Here is a little more detail in case you were wondering about why I picked these roots in particular. As an alternative you could have picked $$-\infty which in the tangent half-angle parametrization corresponds to a rotation by $180°$. This explains why you see three real roots: two describing opposite directions with opposite signs, and one which has its opposite at infinity. So both the ranges above describe exactly the same formula, once with parameter $t$ and once with $-\frac1t$.
First of all from $x^3+y^2=2xy$ you can solve for $y$ to find $$ y=x(1\pm\sqrt{1-x}), $$ so that $0< x\le 1$. A convenient parametrization could then be $$x=\sin^2t \quad\hbox{and}\quad y=\sin^2t(1+\cos t), \quad\hbox{with}\quad 0< t<\pi.$$