A function $y=f\left(x\right)$  satisfies $x{f}^{\mathrm{\prime }}\left(x\right)-2f\left(x\right)=x^4{f}^2\left(x\right),\mathrm{\forall }x>0$ and $f\left(1\right)=-6$ .The value of $f^{\mathrm{\prime }}\left(3^{1/5}\right)$  is

Given , $\frac{xdy}{dx}-2y=x^4{y}^2\text{ or }\frac{dy}{dx}-\frac{2}{x}y=x^3{y}^2$

$\text{ Now dividing by }{y}^2\text{, we get }\frac{1}{y^2}\frac{dy}{dx}-\frac{2}{y}\frac{1}{x}=x^3$

$\text{ This is Bernoulli's differential equation, substituting }\frac{-2}{y}=t\text{, we get }$

$I.F=e^{\int \frac{2}{x}dx}=e^{2\ln x}=e^{\ln x^2}=x^2$

$$\begin{gathered} \text{ So, general solution is given by }{x}^{2}t=\frac{2x^6}{6}+C\Rightarrow \frac{-x^2\cdot 2}{y}=\frac{x^6}{3}+C \\ \Rightarrow \frac{-2}{y}=\frac{x^4}{3}+\frac{C}{x^2} \end{gathered}$$

$\text{ If }x=1,y=-6\text{, then }C=0$

$\frac{-2}{y}=\frac{x^4}{3}\Rightarrow y=\frac{-6}{x^4}\text{ i.e, }f\left(x\right)=\frac{-6}{x^4}$

$\text{ Now, }\frac{dy}{dx}=24x^{-5}=\frac{24}{x^5}$

$\text{ Hence, }{\left.\frac{dy}{dx}\right]}_{x=3^{\frac{1}{5}}}=\frac{24}{3}=8$