Chow test of parameter constancy
This should help you figure out how to compute RSS for the different models.

Let's begin with what we have:

We know that


$y$ is an $n \times 1$ vector that has observations of the endogenous variable
$x$ is an $n \times 2$ matrix of observations of the exogenous variables
$\beta$ is a $2 \times 1$ matrix of coefficients


Unfortunately (or fortunately depending on how you look at it), we don't know what $x$ and $y$ look like in this equation. Instead we are given $(x' x)$, $(x' y)$, and $(y' y)$. 

Using your comment about what RSS actually is -- Namely, that

$$\text{RSS} = (y_{\text{observed}} - y_{\text{estimated}})' (y_{\text{observed}} - y_{\text{estimated}})$$

First notice that $y_{\text{estimated}}$ is simply $x \beta$. Then replacing this in the previous formula gives us

\begin{align*}
  \text{RSS} &amp;= (y_{\text{observed}} - y_{\text{estimated}})' (y_{\text{observed}} - y_{\text{estimated}}) \\
  &amp;= (y_{\text{observed}} - x \beta)' (y_{\text{observed}} - x \beta) \\
  &amp;= y'y - 2y'x\beta + x \beta \beta' x'
\end{align*}

Now this is almost what we want. We want things to ultimately be in terms of $(y'y)$, $(x'x)$, $(x' y)$, and $\beta$. The first term is exactly $y'y$, but other terms look like they have pieces we want and we just have to find a way to get them.

Now, notice that each of the elements in our equation above is simply a scalar (aka sizes are such that they are $1 \times 1$). The determinant of a scalar is itself, then by using properties of determinants we can say:

\begin{align*}
  2y'x \beta &amp;= \det(2y'x\beta) = \det(2 \beta' x' y) \\
  x \beta \beta' x' &amp;= \det(x \beta \beta' x') = \det(\beta' x' x \beta)
\end{align*}

I will leave the linear algebra itself as an exercise.