$f(x){f}''(x)+{{({f}'(x))}^{2}}=(f(x).{f}'(x){)}'$  $f(x)=\frac{1}{\sqrt{x}}\Rightarrow {f}'(x)=\frac{-\frac{1}{2\sqrt{x}}}{{{(\sqrt{x})}^{2}}}\Rightarrow {f}'(x)=\frac{-1}{2x\sqrt{x}}$  $f(x)\times {f}'(x)=\frac{1}{\sqrt{x}}\times \frac{-1}{2x\sqrt{x}}=\frac{-1}{2{{x}^{2}}}$  $f(x)\times {f}'(x)=\frac{-1}{2{{x}^{2}}}\Rightarrow (f(x)\times {f}'(x){)}'=\frac{4x}{4{{x}^{4}}}=\frac{1}{{{x}^{3}}}$  $\frac{1}{{{x}^{3}}}=\frac{k}{{{x}^{n}}}\Rightarrow k=1\,\,,\,\,n=3\Rightarrow k+n=4$