The differential equation having the general solution $y=c{(x-c)}^2$ (c is an arbitrary constant) is 

$$\begin{gathered} y=c{(x-c)}^2 \\ {y}_1=2c(x-c)\Rightarrow x-c=\frac{y_1}{2c} \\ y=c{\left(\frac{y_1}{2c}\right)}^2\Rightarrow c=\frac{y_1^2}{4y} \\ y=c{(x-1)}^2 \\ y\frac{y_1^2}{4y}{(x-\frac{y_1^2}{4y})}^2 \\ 64y^4=y_1^4{(4xy-y_1^2)}^2 \\ 8y^2=y_1(4xy-y_1^2) \\ \therefore y_1^3=4y(x{y}_1-2y) \end{gathered}$$