Denote \(\dfrac{x}{3}=\dfrac{y}{4}=k\), then x = 3k ; y = 4k

We have : \(3k.4k=108\Leftrightarrow12k^2=108\Leftrightarrow k^2=9\Leftrightarrow k=\pm3\)

So, (x ; y) = (9 ; 12) ; (-9 ; -12)

We have:\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{xy}{12}=\dfrac{x^2}{9}=\dfrac{108}{12}=9\Rightarrow x^2=81\)

\(\Rightarrow x=\pm9\)\(\Rightarrow y=\pm12\)

So \(\left(x,y\right)\in\left\{\left(-9,-12\right);\left(9,12\right)\right\}\) satisfy problem