What is the value of $x*y$? 
Given that $$\left(\frac{x}{y}\right)^{-2} + \left(\frac{y}{x}\right)^{-2} = \frac{10}{3}$$
  find the value of $x*y$.

My question is, can we calculate the value of $x*y$ or not? If yes, then how? If not, then why?
 A: As @Meelo said, if (x,y) is a solution so is (cx,cy) when c is not equal to zero.
$$\frac {y^2}{x^2} + \frac {x^2}{y^2} = \frac {y^4+x^4}{x^2y^2}$$
$$\frac{(cy)^2}{(cx)^2}+\frac{(cx)^2}{(cy)^2} = \frac{c^4(y^4+x^4)}{c^4x^2y^2} = \frac{x^4+y^4}{x^2y^2}$$
$$cx\cdot cy = c^2xy$$
Thus we can not find any single value for $xy$ because there are infinite solutions.
A: I think you can using the following result:
$(a^2 + b^2) = (a+b)^2 - 2ab$
$$\left( \frac{x}{y}\right)^{-2} + \left(\frac{x}{y}\right)^{2} = \frac{10}{3} $$
$$\left(\frac{x}{y} + \frac{1}{\frac{x}{y}}\right)^2 - 2\times\frac{x}{y} \times \frac{1}{\frac{x}{y}} = \frac{10}{3}$$
After simplifying you obtain:
$\left(\frac{x^2 + y^2}{xy}\right)^2 = \frac{16}{3} $
Eventually, you get something thats looks like:
$$ \frac{x^2 + y^2}{xy} = \frac{4}{\sqrt3}$$
As pointed out in the other answers, there seems to be infinitely many solutions.
A: no, you really can't. You can find, say, $r = x/y,$ it is one of $\pm \sqrt 3, \pm 1/\sqrt3.$ However, given any $(x,y)$ pair that works, another pair $(tx,ty)$ works, for arbitrary $t.$ there is no restriction on $xy$
A: Some thoughts about this problem:
As observed by @Meelo, the set of solutions contain all non-zero multiples of a given solution, and one  cannot hope to compute $xy$. Actually, the equation  is equivalent to
$$x^4+y^4-\frac{10}3x^2y^2=0, \quad x,y\neq 0$$
which is the equation of a cone in $\mathbf R^2$ – or the equation of a finite number of points on the projective line $\mathbf P_1(\mathbf R)$.
To solve the equation, set $y=tx$. We obtain the biquadratic equation: $\,3t^4-10t^2+3=0$.
It has real roots, since its reduced discriminant is $\,\Delta'=25-9=16$, so that:
$$t^2=\frac{5\pm4}3=\begin{cases}3\\\dfrac13\end{cases},\enspace\text{whence}\quad t=\begin{cases}\pm\sqrt3\\\pm\dfrac1{\sqrt3}\end{cases}$$
A: Let $u = (\frac{x}{y})^2$.
Then, your equation becomes: $\frac{1}{u} + u = \frac{10}{3}$.
Multiplying both sides by u, we get:
$1 + u^2 = \frac{10}{3} \cdot u$
Now, we have a quadratic equation to solve. Subtract $\frac{10}{3} \cdot u$ from both sides:
$1 + u^2 - \frac{10}{3} \cdot u = 0$
For look's sake, let's put our new polynomial in standard form:
$u^2 - \frac{10}{3} \cdot u + 1 = 0$
Now, solve for u. Notice that there are two solutions. So, because we don't have any indication of which u to use, we can only say that y^2 is either approximately $\frac{x}{1.7676} $ or approximately $\frac{x}{0.5657}.$ Therefore, we cannot calculate xy.
