A tangent drawn to the curve $y = f\left(x\right)$ at $P(x, y)$ cuts
the x-axis and y-axis at $A$ and $B$ respectively such that
$BP : AP = 3 : 1$, given that$f\left(1\right) = 1$ then

Equation of tangent at$P(x, y)$
$Y – y = f\text{'} \left(x\right) (X – x)$

so $A=\left(x-y\frac{1}{f^{\mathrm{\prime }}\left(x\right)},0\right)\text{ and }B=\left(0,y-x{f}^{\mathrm{\prime }}\left(x\right)\right)$

$\begin{array}{l}3:1=BP:AP\\ =\sqrt{x^2+x^2{\left(f^{\mathrm{\prime }}\left(x\right)\right)}^2}:\sqrt{y^2/{\left(f^{\mathrm{\prime }}\left(x\right)\right)}^2}+y^2\\ \frac{9}{1}=\frac{x^2\left(1+{\left(f^{\mathrm{\prime }}\left(x\right)\right)}^2\right)}{y^2\left(1+{\left(f^{\mathrm{\prime }}\left(x\right)\right)}^2\right)}{\left(f^{\mathrm{\prime }}\left(x\right)\right)}^2=\frac{x^2{\left(f^{\mathrm{\prime }}\left(x\right)\right)}^2}{y^2}\\ \Rightarrow 9\frac{y^2}{x^2}=\left(f^{\mathrm{\prime }}(x{)}^2\right)\Rightarrow 3\frac{y}{x}=\pm \frac{dy}{dx}\end{array}$

$\Rightarrow \frac{x^3}{y}=\text{ Const or }{x}^{3}y=C$

Since $f\left(1\right)=1,\text{ so Const }=1$

Thus the curve is either $x^3=y\text{ or }{x}^{3}y=1$

The last curve passes through $(2, 1/8)$