Two of the lines represented by $x^3-6x^{2}y+3x{y}^2+\mathrm{d}{y}^3=0$ are perpendicular for

Let $m_1,m_2,m_3$ be the slopes of the three lines represented by the given equation such that $m_1{m}_2=-1$

We have $m_1{m}_2{m}_3=-\frac{1}{d}$ so that $m_3=\frac{1}{d}$

Since $y=m_{3}x$ i.e. $x=dy$ satisfies the given equation, we get

$\begin{array}{l}{d}^3-6d^2+3d+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ \Rightarrow d\left(d^2-6d+4\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\end{array}$

If $d=0$ the given equation represents the lines x = 0 and $x^2-6xy+3y^2=0$ which are not perpendicular.

$\therefore d\ne 0$ and $d^2-6d+4=0$

$\Rightarrow d=\frac{6\pm \sqrt{36-16}}{2}=3\pm \sqrt{5}$

which gives two real values of d.