If $k$ is a positive number and $f$ is a function such that, for every positive number $x$, $\left[f(x^2+1)\right]^{\sqrt{x}}=k$; 
then, for every positive number $y$, $\left[f(\frac{9+y^2}{y^2})\right]^{\sqrt{\frac{12}{y}}}$ is equal to
$\textbf{(A) }\sqrt{k}\qquad \textbf{(B) }2k\qquad \textbf{(C) }k\sqrt{k}\qquad \textbf{(D) }k^2\qquad  \textbf{(E) }y\sqrt{k}$
We are given that \[[f(x^2 + 1)]^{\sqrt(x)} = k\]
We can rewrite $\frac{9+y^2}{y^2}$ as $\frac{9}{y^2} + 1$
Thus, our function is now
\[[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{12}{y}}} = k\]
\[\Rrightarrow[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y} \cdot 4}} = k\]
\[\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{\sqrt{4}} = (k)^{\sqrt{4}}\]
\[\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{2} = (k)^{2} = k^2\]
\[\boxed{D}\]
~JustinLee2017
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