We have:

\(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}=>\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{y}{3}=>\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{2y}{6}\)

\(=>\dfrac{4}{x}=\dfrac{5-2y}{6}=>x\left(5-2y\right)=4.6=24\)

x\(\in\)N so 24 \(⋮\)5-2y

=> 5-2y\(\in\)Ư(24)={-1; -2; -3; -4; -6; -8; -12; -24; 1; 2; 3; 4; 6; 8; 12; 24}

However y\(\in\)N so 2y\(\ge\)4 =>  5-2y \(\le\)1

From two above we have 5-2y\(\in\){-1; -2; -3; -4; -6; -8; -12; -24; 1}

=> 2y\(\in\){6; 8; 4} (because 2y is the odd)

=> y\(\in\){3; 4; 2}

Change and find x