$\sqrt {xy}$ Is Common In Both Equations

Write them as:

$\begin{cases} \sqrt x\left(\color{#D61F06}{x\sqrt x+y\sqrt y}\right)=336\\ \sqrt y\left(\color{#D61F06}{x\sqrt x+y\sqrt y}\right)=112\end{cases}$

Dividing we get $\small{\dfrac{\sqrt x}{\sqrt y}=3\implies \color{#69047E}{x=9y}}$ . Substituting $\color{#69047E}{x=9y}$ in first equation : $(9y)^2+y\underbrace{\sqrt{9y^2}}_{3y}=336\implies y^2=\dfrac{336}{84}=4 \implies y=2$

$x=9y\implies x=9(2)=18$

$\large \therefore x+y=18+2=\boxed{\color{#007fff}{20}}$

Relevant wiki: System of Equations - Problem Solving - Intermediate

Let $x = ay, a \gt 0$ . Then as $x,y \gt 0$ the system of equations can be written as

$a^{2}y^{2} + y\sqrt{ay^{2}} = 336 \Longrightarrow y^{2}(a^{2} + \sqrt{a}) = 336$ and

$y^{2} + ay\sqrt{ay^{2}} = 112 \Longrightarrow y^{2}(1 + a\sqrt{a}) = 112$ .

From this system we see that

$a^{2} + \sqrt{a} = 3(1 + a\sqrt{a}) \Longrightarrow a^{2} - 3a\sqrt{a} + \sqrt{a} - 3 = 0$

$\Longrightarrow a\sqrt{a}(\sqrt{a} - 3) + (\sqrt{a} - 3) = 0 \Longrightarrow (a\sqrt{a} + 1)(\sqrt{a} - 3) = 0$ .

As $a \gt 0$ we cannot have $a\sqrt{a} = -1$ so we conclude that $\sqrt{a} - 3 = 0 \Longrightarrow a = 9$ .

So $y^{2} = \dfrac{336}{a^{2} + \sqrt{a}} = \dfrac{336}{81 + 3} = 4 \Longrightarrow y = 2 \Longrightarrow x = ay = 9 \times 2 = 18$ ,

and thus $x + y = 18 + 2 = \boxed{20}$ .

Tale $\sqrt {x}$ common from first equation, and $\sqrt{y}$ common from second equation. Divide to get $x=9y$ . Put in the first equation to get $y=2$ and $x=18$