The problem can be simplified by setting 3x = t

 

Then, the statement becomes

 

I₁+I₂ = ₀ f'(x) cosx dx + ₀ f'"(x) cosx dx = 12

 

I₂ = ₀ f'"(x) cosx dx = |f"(x) cosx|₀ +  ₀ f''(x) sinx dx

 

₀ f''(x) sinx dx =  -₀ f'(x) cosx dx (the first part becomes zero at the limits)

 

Hence I₂+I₁ = |f"(x) cosx|₀ = 12

 

-f"() - f"(0) = 12

 

Hence f"(0) = -4-12 = -16

 

PS: You will need magnifying glasses to distinguish f"(x) from f'(x)