KVPY-SA 2017

$\large \displaystyle 72^x 48^y = 6^{xy}$

$\large \displaystyle 2^{3x} \cdot 3^{2x} \cdot 2^{4y} \cdot 3^y = 2^{xy} \cdot 3^{xy}$

$\large \displaystyle 2^{3x+4y} \cdot 3^{2x+y} = 2^{xy} \cdot 3^{xy}$

Since $2$ and $3$ are primes:

$\large \displaystyle \color{#20A900} (i) \ 3x+ 4y = xy$

$\large \displaystyle \color{#20A900} (ii) \ 2x+ y = xy$

Make $(i) - (ii)$ :

$\large \displaystyle \color{#20A900} (iii) \ x = -3y$

Put $(iii)$ on $(i)$ :

$\large \displaystyle -9y + 4y = -3y^2$

Since $y \neq 0$ :

$\color{#20A900} \boxed{ \large \displaystyle y = \frac53}$

Putting this on $(iii)$ :

$\color{#20A900} \boxed{ \large \displaystyle x = -5}$

This:

$\color{#3D99F6} \boxed{ \large \displaystyle x+y = -\frac{10}{3} \approx -3.333}$