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I'm given that $R(y)=p(y) * y$ where $p(y)$ and $R(y)$ are functions of $y$

In the question: I need the product rule to find the derivative of $p(y) \cdot y$

I know that the product rule is $\frac{d}{dx}(f(x)\cdot g(x)) = f'(x)g(x) + g'(x)f(x)$.

Would this then give me $p*y + p(y) * 0$?

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    You should end up with $p'(y)y+p(y)$... – 2017-01-25
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    The derivative of $y \mapsto y$ is $y \mapsto 1$ not $y \mapsto 0$. – 2017-01-25

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$$ \frac{d}{dy}(p(y)\cdot y) = p(y) \frac{d}{dy}(y) + y\frac{d}{dy}(p(x)) \\ \implies \frac{d}{dy}(p(y)\cdot y) = p(y) + y\cdot p'(y). $$

Two things you've mistaken:

$(1)$ The derivative of $y$ with respect to $y$ is $1$ and not $0$. If you define $f(y) = y$, then it is immediately obvious from the definition of a derivative:

$$ \frac{d}{dy}(f(y)) = \lim_{\delta y \to 0} \frac{f(y+\delta y) - f(y)}{\delta y} = \lim_{\delta y\to 0} \frac{y+\delta y-y}{\delta y} = 1. $$

$(2)$ The derivative of $p(y)$ would be denoted by $p'$ and not $p$.