A straight line touches the rectangular hyperbola $9x^2-9y^2=8$ and the parabola $y^2=32x$. An equation of the line is 

Let the equation be $y= mx + c.$ Since it touches the hyperbola 

$9x^2-9y^2=8\text{ or }\frac{x^2}{8/9}-\frac{y^2}{8/9}=1$

$\text{ (with }{a}^2=b^2=8/9\text{ ), }$we get

                                              $c^2=(8/9)\left(m^2-1\right)$            (i)

Also, since it touches the parabola $y^2=32x$, we get

                                          $c=8/m$                               (ii)

From (i) and (ii) we get

           $\frac{8}{9}\left(m^2-1\right)=\frac{64}{m^2}\Rightarrow m^2\left(m^2-1\right)=72$

$\begin{array}{l}\Rightarrow m^4-m^2-72=0\\ \Rightarrow m=3\end{array}\Rightarrow \begin{array}{r}\left(m^2-9\right)\left(m^2+8\right)=0\\ \left[m^2=-8\text{ is rejected }\right]\end{array}$

From (ii), $c=8/3\text{ and }-8/3\text{ when }m=3\text{ and }-3$

respectively. Therefore, the required equations are 

$y=3x+\frac{8}{3}\text{ and }y=-3x-\frac{8}{3}$

or       $9x-3y+8=0\text{ or }9x+3y+8=0$