Answer: \(y=\left(\dfrac{10}{x}+3\right)^3\).

Let

\(X=\frac1x,\qquad Y=\sqrt[3]{y}.\)

Then \(Y\) is a linear function of \(X\).

The line passes through \((0.2,5)\) and \((1,13)\), so its gradient is

\(m=\frac{13-5}{1-0.2}=\frac{8}{0.8}=10.\)

Using the point \((1,13)\),

\(Y-13=10(X-1).\)

So

\(Y=10X+3.\)

Substitute back:

\(\sqrt[3]{y}=10\left(\frac1x\right)+3=\frac{10}{x}+3.\)

Cube both sides:

\(\boxed{y=\left(\frac{10}{x}+3\right)^3}.\)