\(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)(1)

\(\Leftrightarrow\dfrac{xy-12}{3y}=\dfrac{1}{5}\)

\(\Leftrightarrow5xy-60=3y\)

\(\Leftrightarrow5xy-3y=60\)

\(\Leftrightarrow y\cdot\left(5x-3\right)=60\)

\(\Rightarrow y=\dfrac{60}{5x-3}\)(2).

(1),(2) => \(\dfrac{x}{3}-\dfrac{4}{\dfrac{60}{5x-3}}=\dfrac{1}{5}\Leftrightarrow\dfrac{x}{3}-\dfrac{20x-12}{60}=\dfrac{1}{5}\Leftrightarrow\dfrac{x}{3}-\dfrac{4\cdot\left(5x-3\right)}{60}=\dfrac{1}{5}\Leftrightarrow\dfrac{x}{3}-\dfrac{5x-3}{15}=\dfrac{1}{5}\Leftrightarrow\dfrac{5x}{15}-\dfrac{5x-3}{15}=\dfrac{1}{5}\Leftrightarrow\dfrac{3}{15}=\dfrac{1}{5}\) . 

From here we know that x has lots of value but just (x;y) = (1;30),(3;5) satisfy thread.

Dao Trong Luan's solution is more logical.You should choose his.

x3−4y=15x3−4y=15(1)

⇔xy−123y=15⇔xy−123y=15

⇔5xy−60=3y⇔5xy−60=3y

⇔5xy−3y=60⇔5xy−3y=60

⇔y⋅(5x−3)=60⇔y⋅(5x−3)=60

⇒y=605x−3⇒y=605x−3(2).

(1),(2) => x3−4605x−3=15⇔x3−20x−1260=15⇔x3−4⋅(5x−3)60=15⇔x3−5x−315=15⇔5x15−5x−315=15⇔315=15x3−4605x−3=15⇔x3−20x−1260=15⇔x3−4⋅(5x−3)60=15⇔x3−5x−315=15⇔5x15−5x−315=15⇔315=15 . 

From here we know that x has lots of value but just (x;y) = (1;30),(3;5) satisfy thread.

\(\Rightarrow\dfrac{4}{y}=\dfrac{x}{3}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{5x}{15}-\dfrac{3}{15}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{5x-3}{15}\)

\(\Rightarrow y\left(5x-3\right)=60\)

Because \(5x-3\equiv2\left(mod5\right)\) and 5x-3 \(\in U\left(60\right)\)

\(\Rightarrow5x-3\in\left\{2;12\right\}\)

If 5x - 3 = 2

=> 5x = 5 => x = 1

And y = 60: 2 = 30

If 5x  - 3 = 12

=> 5x = 15 => x = 3

And y = 60 : 12 = 5

So \(\left(x,y\right)=\left(1;30\right),\left(3;5\right)\)