$\text{ Let }f:R\to R\text{ be a continuous and differentiable function such that }{\left(f\left(x^2+1\right)\right)}^{\sqrt{x}}=5\text{ for }\mathrm{\forall }x\in (0,\mathrm{\infty })\text{ . Then the value }$

$\text{ of }{\left(f\left(\frac{16+y^2}{y^2}\right)\right)}^{\frac{4}{\sqrt{y}}}\text{ for }y\in (0,\mathrm{\infty })\text{ is equal to }$

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