Answer: \(x=\log_8\!\left(\frac34\right)\), approximately \(-0.138\).

Rewrite both exponential terms using base \(2\):

\(64^{x+\frac13}=(2^6)^{x+\frac13}=2^{6x+2}.\)

Also

\(2^{3x}=8^x.\)

Let

\(u=2^{3x}=8^x.\)

Then

\(2^{6x+2}=4(2^{3x})^2=4u^2.\)

So the equation becomes

\(4u^2+u-3=0.\)

Factorise:

\((4u-3)(u+1)=0.\)

Since \(u=2^{3x}\gt 0\), the value \(u=-1\) is impossible. Therefore

\(u=\frac34.\)

So

\(8^x=\frac34.\)

Hence

\(x=\log_8\left(\frac34\right).\)

Numerically,

\(x=-0.138\ldots\)