Subtracting the second equation from the first and multiplying by $xy$ we get $x^2y + y - xy^2 - x = 0$, which gives either $x = y$ or $x = \frac{1}{y}$.
If $x = y$, then subtracting the fourth equation from the third and multiplying by $xz$, we similarly get either $x = z$ or $x = -\frac{1}{z}$. If $x = z$, then subtracting the third equation from the first gives $-2w = 0$, hence $w = 0$, which is not positive. If $x = -\frac{1}{z}$, then $x$ and $z$ cannot be both positive.
If $x = \frac{1}{y}$, then subtracting the fourth equation from the third gives $z - \frac{1}{z} = 0$, hence $z = 1$. Adding the first equation with the third gives $x + \frac{2}{x} = 3$, or $x^2 - 3x + 2 = 0$. This has a solution $x = 1$, which leads to $x = y = z$ already considered. The second solution $x = 2$ gives $y = \frac{1}{2}$ and $w = \frac{1}{2}$.