The tangent line to [tex]y(x)[/tex] at [tex]x=\dfrac\pi4[/tex] has slope [tex]y'\left(\dfrac\pi4\right)[/tex], so first we compute the derivative:
[tex]y'(x)=9\sec^2x[/tex]
and evaluate it at the point of interest to find the slope:
[tex]y'\left(\dfrac\pi4\right)=9\sec^2\dfrac\pi4=18[/tex]
Then the equation of the tangent line is
[tex]y-9=18\left(x-\dfrac\pi4\right)\implies y=18x+9-\dfrac{9\pi}2[/tex]
