Answer
$\mathop \smallint \limits_{y = 0}^9 \mathop \smallint \limits_{x = 0}^{\sqrt y } \frac{x}{{{{\left( {{x^2} + y} \right)}^{1/2}}}}{\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = {x^2}}^9 \frac{x}{{{{\left( {{x^2} + y} \right)}^{1/2}}}}{\rm{d}}y{\rm{d}}x \simeq 7.46$
Work Step by Step
From the integration $\mathop \smallint \limits_{y = 0}^9 \mathop \smallint \limits_{x = 0}^{\sqrt y } \frac{{x{\rm{d}}x{\rm{d}}y}}{{{{\left( {{x^2} + y} \right)}^{1/2}}}}$ we obtain the domain description:
$0 \le x \le \sqrt y $, ${\ \ \ }$ $0 \le y \le 9$
Notice that this is a horizontally simple region.
We change the region to vertically simple region such that the order of integration becomes $dydx$. Using $x = \sqrt y $ and $y=9$, we obtain the boundaries $y = {x^2}$ and $x=3$. This is illustrated in the figure attached.
We see that the region is lower bounded by $y = {x^2}$ and upper bounded by $y=9$. Thus, the description of the domain as a vertically simple region is
$0 \le x \le 3$, ${\ \ \ }$ ${x^2} \le y \le 9$
Evaluate the integral in the order of $dydx$:
$\mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = {x^2}}^9 \frac{x}{{{{\left( {{x^2} + y} \right)}^{1/2}}}}{\rm{d}}y{\rm{d}}x = 2\mathop \smallint \limits_{x = 0}^3 x\left( {{{\left( {{x^2} + y} \right)}^{1/2}}|_{{x^2}}^9} \right){\rm{d}}x$
$ = 2\mathop \smallint \limits_{x = 0}^3 x\left( {{{\left( {{x^2} + 9} \right)}^{1/2}} - \sqrt 2 x} \right){\rm{d}}x$
$ = 2\left( {\frac{1}{3}{{\left( {{x^2} + 9} \right)}^{3/2}}|_0^3 - \frac{{\sqrt 2 }}{3}{x^3}|_0^3} \right)$
$ = 2\left( {\frac{1}{3}\left( {54\sqrt 2 - 27} \right) - 9\sqrt 2 } \right) = 18\left( {\sqrt 2 - 1} \right) \simeq 7.46$
So, $\mathop \smallint \limits_{y = 0}^9 \mathop \smallint \limits_{x = 0}^{\sqrt y } \frac{x}{{{{\left( {{x^2} + y} \right)}^{1/2}}}}{\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = {x^2}}^9 \frac{x}{{{{\left( {{x^2} + y} \right)}^{1/2}}}}{\rm{d}}y{\rm{d}}x \simeq 7.46$.
