Answer: \(y=\frac{3}{2}(x^2-1)\).

Let \(y=vx\). Then

\(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\).

Substitute into the differential equation:

\(x\left(v+x\frac{dv}{dx}\right)=vx+\sqrt{9x^2+y^2}\).

Since \(y=vx\), this becomes

\(x\left(v+x\frac{dv}{dx}\right)=vx+\sqrt{9x^2+v^2x^2}\).

So

\(xv+x^2\frac{dv}{dx}=xv+\sqrt{x^2(v^2+9)}\).

With \(x=1\) in the given condition, we work on the branch \(x\gt 0\), so \(\sqrt{x^2}=x\). Hence

\(x^2\frac{dv}{dx}=x\sqrt{v^2+9}\),

so

\(x\frac{dv}{dx}=\sqrt{v^2+9}\).

Separate variables:

\(\displaystyle \frac{dv}{\sqrt{v^2+9}}=\frac{dx}{x}\).

Integrating gives

\(\displaystyle \sinh^{-1}\left(\frac{v}{3}\right)=\ln x+C\).

Now use \(y=0\) when \(x=1\). Since \(y=vx\), at \(x=1\) we have \(v=0\). Therefore

\(\sinh^{-1}(0)=\ln 1+C\),

so \(C=0\).

Thus

\(\displaystyle \sinh^{-1}\left(\frac{v}{3}\right)=\ln x\).

Write this in logarithmic form:

\(\displaystyle \ln\left(\frac{v}{3}+\sqrt{\frac{v^2}{9}+1}\right)=\ln x\).

Hence

\(\displaystyle \frac{v}{3}+\sqrt{\frac{v^2}{9}+1}=x\).

Substitute \(v=\frac{y}{x}\) and multiply through by \(3x\):

\(y+\sqrt{y^2+9x^2}=3x^2\).

Now square both sides:

\((3x^2-y)^2=y^2+9x^2\).

Expanding,

\(9x^4-6x^2y+y^2=y^2+9x^2\).

So

\(9x^4-6x^2y-9x^2=0\).

Hence

\(6x^2y=9x^4-9x^2\),

and therefore

\(\displaystyle y=\frac{3}{2}(x^2-1)\).