The equation of the curve which passes through point \((1,0)\) and has tangent with
slope \(1+\frac{y}{x}+\left(\frac{y}{x}\right)^{2}\) is

We have siope of tangent \(=\frac{d y}{d x}=1+\frac{y}{x}+\left(\frac{y}{x}\right)^{2}\)
\(\therefore \frac{d y}{d x}=1+\frac{y}{x}+\frac{y^{2}}{x^{2}} \Rightarrow \frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}}\) ...(1)
Put \(y=u x \Rightarrow \frac{d y}{d x}=u+x \frac{d u}{d x}\)
\(\therefore u+x \frac{d u}{d x}=\frac{x^{2}+u x^{2}+u^{2} x^{2}}{x^{2}}\)
\(\therefore u+x \frac{d u}{d x}=1+u+u^{2} \Rightarrow x \frac{d u}{d x}=1+u^{2} \Rightarrow\) \(\int \frac{d u}{1+u^{2}}=\int \frac{d x}{x}\)
\(\therefore \tan ^{-1} u=\log |x|+c \Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\log |x|+c\)
At \((1,0)\), we write \(\tan ^{-1}(0)=\log |1|+c \Rightarrow c=0\)
\(\therefore \tan ^{-1}\left(\frac{y}{x}\right)=\log |x|\)