Find $xy$

Consider

$\begin{aligned} \left(x + \frac 1x\right)^4 & = x^4 + 4x^2 + 6 + \frac 4{x^2} + \frac 1{x^4} \\ & = {\color{#3D99F6} x^4 + \frac 1{x^4} + 4\left(x^2+\frac 1{x^2} \right)} + 6 \\ & = {\color{#3D99F6}75} + 6 \\ & = 81 \\ \implies x + \frac 1x & = 3 \quad \quad \small \color{#3D99F6} \text{For }x > 0 \\ x + y & = 3 \quad \quad \small \color{#3D99F6} \text{Given} \\ \implies y & = \frac 1x \\ \implies xy & = x \cdot \frac 1x = \boxed{1} \end{aligned}$

$x^{4}+x^{-4}+4(x^{2}+x^{-2})=75\\ \text{Adding 6 to both sides,}\\ x^{4}+\frac{1}{x^{4}}+4(x^{2}+\frac{1}{x^{2}})+6=81\\(x+\frac{1}{x})^{4}=81\\ x+\frac{1}{x}=±3 \\ \text{We will take x=3 as x>0}\\ \implies{\frac{1}{x}=y}\\ \therefore{xy=x\times{\frac{1}{x}}=\boxed{1}}$