Answer:

[tex]\boxed{ \sf{ \:\bf \:(2) \: \: \: x = \: \dfrac{k \: + \: \sqrt{k}}{3} \: \: or \: \: x = \: \dfrac{k \: - \: \sqrt{k}}{3} \: }}\\ \\ [/tex]

Step-by-step explanation:

Given curve is

[tex]\sf \: y=k(3x-k) ^{-1}+3x \\ \\ [/tex]

On differentiating both sides w. r. t. x, we get

[tex]\sf \: \dfrac{d}{dx}y=\dfrac{d}{dx}[k(3x-k) ^{-1}+3x] \\ \\ [/tex]

[tex]\sf \: \dfrac{dy}{dx}=k\dfrac{d}{dx}(3x-k) ^{-1}+3\dfrac{d}{dx}x \\ \\ [/tex]

We know,

[tex] \qquad \:\boxed{\begin{aligned}&\sf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \\ \\&\sf \: \dfrac{d}{dx}x=1\\\end{aligned}} \\ \\ [/tex]

So, using these results, we get

[tex]\sf \: \dfrac{dy}{dx}= - k {(3x - k)}^{ - 1 - 1} \dfrac{d}{dx}(3x-k)+3 \times 1 \\ \\ [/tex]

[tex]\sf \: \dfrac{dy}{dx}= - k {(3x - k)}^{ -2} (3) + 3 \\ \\ [/tex]

[tex]\sf \: \dfrac{dy}{dx}= - 3k {(3x - k)}^{ -2} + 3 \\ \\ [/tex]

For stationary points,

[tex]\sf \: \dfrac{dy}{dx}= 0 \\ \\ [/tex]

[tex]\sf \: - 3k {(3x - k)}^{ -2} + 3 = 0 \\ \\ [/tex]

[tex]\sf \: - 3k {(3x - k)}^{ -2} = - 3\\ \\ [/tex]

[tex]\sf \:k {(3x - k)}^{ -2} = 1\\ \\ [/tex]

[tex]\sf \:{(3x - k)}^{2} = k\\ \\ [/tex]

[tex]\sf \: 3x - k = \: \pm \: \sqrt{k} \\ \\ [/tex]

[tex]\sf \: 3x = \: k \: \pm \: \sqrt{k} \\ \\ [/tex]

[tex]\sf \: x = \: \dfrac{k \: \pm \: \sqrt{k}}{3} \\ \\ [/tex]

[tex]\bf\implies \:x = \: \dfrac{k \: + \: \sqrt{k}}{3} \: \: or \: \: x = \: \dfrac{k \: - \: \sqrt{k}}{3} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Stationary points are defined as those points on the curve at which derivative is 0 or those points on the curve at which tangent is parallel to x - axis.