Ok, so I have function $$f(x)=\dfrac{x^3+6x^2-3x}{x^2+1}$$ All answers have to be completed on maxima. The first part is to plot the graph, which I have. The second part is to find the derivative of $f$. I have also completed that.
Part C asks "calculate the $x$ and $y$ coordinates of the local maximum of $f$.
I know I need to find $f'(x)=0$ but no matter what I do, I just cannot get what I'm looking for!
I agree that you face a quite unpleasant problem since $$f(x)=\dfrac{x^3+6x^2-3x}{x^2+1}\implies f'(x)=\dfrac{x^4+6x^2+12x-3}{(x^2+1)^2}$$ So, you need to solve the quartic $$x^4+6x^2+12x-3=0$$ If you look here, the method is described.
The first point is $\Delta=-1168128$ which means that the equation has two distinct real roots and two complex conjugate non-real roots (this is good news since you now know that there is only one maximum and one minimum).
The bad news is that, using the method, the real roots are given by the ugly $$x_{\pm}=\frac{1}{\sqrt{\frac{2}{-4-\sqrt[3]{26}+6 \sqrt{\frac{2}{\sqrt[3]{26}-2}}}}}\pm\sqrt{\frac{1}{2} \left(\sqrt[3]{26}-2\right)}$$ which are $x_-\approx -1.61201$ and $x_+\approx 0.224572$. Use these last numbers and your pocket calculator to get $f(x_-)\approx 4.51$ and $f(x_+)\approx -0.34$.