General solution of the differential equation $y=x\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\mathrm{d}x}{\mathrm{d}y}$represent 

$y=x\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\mathrm{d}x}{\mathrm{d}y}$

Putting $p=\frac{\mathrm{d}y}{\mathrm{d}x}$, the given equation can be written as $y=px+\frac{1}{p}$

On differentiating w.r.t. x, we have 

$$\begin{gathered} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}=p+x\frac{\mathrm{d}p}{\mathrm{d}x}-\frac{1}{p^2}\frac{\mathrm{d}p}{\mathrm{d}x} \\ \Rightarrow \left(x-\left(1/p^2\right)\right)\frac{\mathrm{d}p}{\mathrm{d}x}=0 \\ \Rightarrow p^2=\frac{1}{x}\text{ or }\frac{\mathrm{d}p}{\mathrm{d}x}=0 \\ \frac{\mathrm{d}p}{\mathrm{d}x}=0\text{ then }p=\mathrm{constant}=c \end{gathered}$$

putting this value in given equation, we get $y=cx+1/c$ which represents a straight line 

If $p^2=1/x\text{ then }{y}^2=(px+1/p{)}^2=p^2{x}^2+1/p^2 +2x=\frac{1}{x}{x}^2+x+2x=4x$ which represents a parabola.