I don't know all those but I do know how to do (d).
\(\dfrac{dy}{dx}\\=\dfrac{d}{dx} \left(\ln \left(x^4\right)+e^x\right)\\=\dfrac{d}{dx}(4\ln x)+\dfrac{d}{dx}(e^x)\)
\(=4\dfrac{d}{dx}(\ln x)+\dfrac{d}{dx} e^x\)
\(=\dfrac{4}x+e^x\)
\(\begin{array}{rl}z_0=e^u&u=2x^2\\\end{array}\\ \dfrac{dz_0}{dx}=\dfrac{dz_0}{du}\times \dfrac{du}{dx}\leftarrow\text{ Chain rule}\\ \;\;\;\;\;\;\!=e^u\times 4x\\ \;\;\;\;\;\;\!=4xe^{2x^2}\)
You can see that I am differentiating z0 =e^2x^2 using chain rule.
\(\dfrac{dz}{dx}\\ =\dfrac{d}{dx}(z_0)+\dfrac{d}{dx}(6)\\ =4xe^{2x^2}\)
\(\therefore\dfrac{dp}{dx}=\dfrac{dy}{dx}+\dfrac{dz}{dx}=\dfrac{4}{x}+e^x+4xe^{2x^2}\)
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