Let  $\mathrm{y}=\mathrm{y}\left(\mathrm{x}\right)$ be the solution curve of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}(1+{\mathrm{xy}}^2(1+{\log }_{\mathrm{e}}\mathrm{x})),\mathrm{\hspace{0.17em}}\mathrm{x}>0,\mathrm{\hspace{0.17em}}\mathrm{y}\left(1\right)=3$ . Then  $\frac{{\mathrm{y}}^2\left(\mathrm{x}\right)}{9}$ is equal to: 

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}={\mathrm{y}}^3(1+\mathrm{lnx})\\ \Rightarrow \frac{1}{{\mathrm{y}}^3}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{{\mathrm{xy}}^2}=(1+\mathrm{lnx})\\ \mathrm{Taking}\frac{1}{{\mathrm{y}}^2}=\mathrm{t}\\ \Rightarrow -\frac{2}{{\mathrm{y}}^3}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}\\ \therefore -\frac{1}{2}\frac{\mathrm{dt}}{\mathrm{dx}}-\frac{\mathrm{t}}{\mathrm{x}}=(1+\mathrm{lnx})\\ \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}+\frac{2\mathrm{t}}{\mathrm{x}}=-2(1+\mathrm{lnx})\\ \mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \frac{2}{\mathrm{x}}\mathrm{dx}}={\mathrm{x}}^2\\ \therefore {\mathrm{tx}}^2=\int -2(1+\mathrm{lnx}){\mathrm{x}}^2\mathrm{dx}\\ \Rightarrow {\mathrm{tx}}^2=-2[\frac{(1+\mathrm{lnx}){\mathrm{x}}^3}{3}-\int \frac{{\mathrm{x}}^2}{3}\mathrm{dx}]+\mathrm{c}\\ \frac{{\mathrm{x}}^2}{{\mathrm{y}}^2}=-2[\frac{{\mathrm{x}}^3}{3}(1+\mathrm{lnx})-\frac{{\mathrm{x}}^3}{9}]+\mathrm{c}........\left(\mathrm{i}\right)\\ \mathrm{y}\left(1\right)=3\Rightarrow \frac{1}{9}=-2(\frac{1}{3}-\frac{1}{9})+\mathrm{c}\\ \therefore \mathrm{c}=\frac{5}{9}\end{array}$
$\begin{array}{l}\therefore \frac{x^2}{y^2}=-2\left(\frac{x^3}{3}(1+\ln x)-\frac{x^3}{9}\right]+\frac{5}{9}\\ \Rightarrow \frac{y^2}{9}=\frac{x^2}{5-2x^3\left(2+\ln x^3\right)}\end{array}$

Explanation:

To solve the given differential equation dydx=yx(1+xy2(1+lnx)) with the initial condition y(1)=3, we will follow these steps:

Step 1: Rearranging the Differential Equation
We start with the equation:
dydx=yx(1+xy2(1+lnx))
We can rewrite this as:
dydx=yx+xy3(1+lnx)x
This simplifies to:
dydx=yx+y3(1+lnx)

Step 2: Separating Variables
We can separate the variables:
dyy3=(1+lnx)dx−1xdy
Rearranging gives us:
dyy3=(1+lnx−1x)dx

Step 3: Integrating Both Sides
Now we integrate both sides:
∫dyy3=∫(1+lnx−1x)dx
The left side integrates to:
−12y2
The right side can be integrated term by term:
∫1dx+∫lnxdx−∫1xdx=x+xlnx−x+C
Thus, we have:
−12y2=xlnx+C

Step 4: Applying the Initial Condition
Using the initial condition y(1)=3:
−12(32)=1⋅ln(1)+C
This simplifies to:
−118=0+C⟹C=−118

Step 5: Substituting Back
Substituting C back into our equation gives:
−12y2=xlnx−118
Multiplying through by -2:
1y2=−2xlnx+19

Step 6: Finding y29
Rearranging gives:
y2=91−18xlnx
Thus, we find:
y29=11−18xlnx

Final Answer
The value of y2(x)9 is:
y2(x)9=11−18xlnx