<s>[INST] Finding the minimum of $x^2+y^2$ for $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$ and $xy>0$. If $x,y \in \mathbb {R}$, find the minimum of  $x^2+y^2$ when $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$ and $xy>0$. 
This problem was inspired by a problem which asked if $x,y \in \mathbb {R}$ and $xy \neq 0$,  find the minimum of $x^2+y^2$ when $xy(x^2-y^2)=x^2+y^2$.
By setting $x=a\sin\theta$, $y=a\cos\theta$, the equation can be simplified to $a^2=\frac{1}{\sin\theta\cos\theta(\sin\theta^2-\cos\theta^2)}$
However, notice that $\sin\theta\cos\theta=\frac{\sin2\theta}{2}$, $\sin\theta^2-\cos\theta^2=-\cos2\theta$. 
This implies that $a^2=\frac{2}{-\sin2\theta\cos2\theta}$, thus that $a^2=\frac{4}{-\sin4\theta}\ge 4$. 
Thus the minimum of $x^2+y^2$ is $4$, with the equality holding when $x=\sqrt{2-\sqrt{2}}$, $y=\sqrt{2+\sqrt{2}}$.
However, since there are no formulas I know of where $\sin^3 x+\cos^3 x$, I did not know how to find the minimum of $x^2+y^2$ when $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$.
Graphing it seems to imply that such a minimum exists, but I am not aware of how to find it( and when such a minimum exists). 
Any help would be appreciated. 
 [/INST] EDIT: Just realized $xy≠0$. Here is the sadly tough solution.
Instead of writing out the minutiae, I will just sketch out the solution method. 
To begin, rewrite the initial expression as follows: 
$xy(x-y)^2(x^2+xy+y^2)=(x+y)(x^2+y^2-xy)$
We did this by factoring the sum and the difference of cubes. (For which there are established formulas, but which also may be derived). 
Now, simply make the substitutions you used, though I would call $x=r\cos(\theta)$ and $y=r\sin(\theta)$ because of the Pythagorean definitions of the trigonometric functions. ("Opposite" the angle corresponding to the $y$ length and "adjacent" corresponding to the abscissa ($x$-axis) length)
You will, solving for $r$ or $a$ or whatever you call it, find an expression in $\theta$ for the cube of that number. Letting $r=\sqrt{x^2+y^2}$, we find that: 
$r^3=\frac{2(\sin(\theta)+\cos(\theta))(2-\sin(2\theta))}{(sin(2\theta))(1-\sin(‌​\theta))(2+\sin(2\theta))}$
And that, 
$r^3 = -\frac{2(\sin(2x)-2)\cot(2x)}{(\sin(2x)+2)(\cos(x)-\sin(x))^3}$
Take the cube root of both sides, square booth sides, then find the minimum of the resulting expression in $\theta$ on the RHS. 
This will be the minimum value. Writing the last sentence was much easier than doing out the math. Because trig functions are periodic, if you consider an interval $0<\theta<2π$ you need to locate the relative minima, of which it becomes apparent that $0$ is the lowest function value taken. 
In a world without WolframAlpha we could use some serious single-variable calculus to determine the relative minima in the given interval by locating where the function's derivatives are zero and applying the usual test. 
However, you say that $xy≠0$. Aww, man. But you are right, that would have been too easy. 
We now want to look at the other relative minima. Computationally, the answer is 3.182. I do not believe there are ways to compute it by hand.
$r=\left(-\frac{2\left(\sin \left(2\theta\right)-2\right)\cot \left(2\theta\right)}{\left(\sin \left(2\theta\right)+2\right)\left(\cos \left(\theta\right)-\sin \left(\theta\right)\right)^3}\right)^{\frac{2}{3}}$
You want the minima of that function that is nonzero with x and y greater than zero in the original equation.
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