The given inequality can be transformed to various equivalent forms: $$9^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^9 \iff (3^2)^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^{3\times 3} \iff 3^{2\sqrt{2}} \stackrel{?}{<} {(2\sqrt{2})}^3 \iff 3^{\frac13} \stackrel{?}{<} {(2\sqrt{2})}^{\frac{1}{2\sqrt{2}}}$$$$\begin{align} 9^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^9 &\iff (3^2)^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^{3\times 3} \iff 3^{2\sqrt{2}} \stackrel{?}{<} {(2\sqrt{2})}^3\\ &\iff 3^{\frac13} \stackrel{?}{<} {(2\sqrt{2})}^{\frac{1}{2\sqrt{2}}}\end{align}$$
Notice $3 > 2\sqrt{2} \sim 2.828 > e ~ 2.718 $$3 > 2\sqrt{2} \sim 2.828 > e \sim 2.718 $ and the function $x^{\frac1x}$ is strictly decreasing for $x > e$ (standard calculus exercise). The rightmost "inequality" is true and hence the original "inequality" $9^{\sqrt{2}} < \sqrt{2}^9$ is true.