If the curve $y=f\left(x\right)$ passes through $\left(1,1\right)$ and satisfies the differential equation $2x^{2}ydx-x^{3}dy=y^{2}dx$ .Then$f\left(2\right)$ is equal to

Given , the curve $y=f\left(x\right)$ passes through $(1,1)$

The given differential equation is  $2x^{2}ydx-x^{3}dy=y^{2}dx$

Separating the terms and integrating them

$$\begin{gathered} \int \frac{2xydx- x^{2}dy}{y^2}=\int \frac{dx}{x} \\ \int d\left(\frac{x^2}{y}\right)=\int \frac{dx}{x}\dots \dots \left(i\right) \end{gathered}$$

As differentiation of $\frac{x^2}{y}= d\left(\frac{x^2}{y}\right)=\frac{2xydx-x^{2}dy}{y^2}$

Since formula of $d\left(\frac{u}{v}\right)=\frac{u\text{'}vdu-uv\text{'}dv}{v^2}$

$$\begin{gathered} \left(i\right)\dots \dots \frac{x^2}{y}=\log \left(x\right)+c \\ \Rightarrow y=\frac{x^2}{\log \left(x\right)+c} \end{gathered}$$

$y=\frac{x^2}{\log x+c}$ passes through $(1,1)$

$\Rightarrow 1=\frac{1^2}{\log \left(1\right)+c}=\frac{1}{c}$ as $\log \left(1\right)$ is equal to zero

$\Rightarrow c=1$

To calculate $f\left(2\right)$ we need to insert 2 in place of x

$$\begin{gathered} f\left(2\right)=\frac{2^2}{\log \left(2\right)+1} \\ f\left(2\right)=\frac{4}{\log \left(2\right)+\log \left(e\right)} \\ f\left(2\right)=\frac{4}{\log \left(2e\right)} \end{gathered}$$

Therefore, the correct answer is option 3.