Answer: \(y=\frac12(3x-1)^{\frac43}+2x-3\).

Integrate once:

\(\frac{dy}{dx}=\int 2(3x-1)^{-\frac23}\,dx.\)

Let \(u=3x-1\), so \(du=3\,dx\). Then

\(\frac{dy}{dx}=2(3x-1)^{\frac13}+c.\)

The gradient is \(6\) when \(x=3\). Since \(3x-1=8\),

\(6=2(8)^{\frac13}+c.\)

So

\(6=4+c,\)

and hence

\(c=2.\)

Therefore

\(\frac{dy}{dx}=2(3x-1)^{\frac13}+2.\)

Integrate again:

\(y=\frac12(3x-1)^{\frac43}+2x+d.\)

The curve passes through \((3,11)\), so

\(11=\frac12(8)^{\frac43}+6+d.\)

Now \(8^{1/3}=2\), so \(8^{4/3}=16\). Hence

\(11=8+6+d,\)

giving

\(d=-3.\)

So the equation of the curve is

\(\boxed{y=\frac12(3x-1)^{\frac43}+2x-3}.\)