What method should I use to find the running time of the recurrence function:
$$T(n)=T(n/4)+T(3n/4)+\theta(n)$$
I've tried to solve this question by guess-and-check. I assumed that the total size of on each level of the tree is less than $n$, so I guessed that $f(n)=n$ will dominate.
How should I go about this?
You will get some intuition from looking at the case 2 example in [1]. Since your first guess was $n$, consider $g(n)=T(n)/n$. This satisfies $$g(n)=\frac{1}{4}g(n/4)+\frac{3}{4}g(3n/4)+\Theta(1) \,.$$ Let's ignore the implicit constant and write this as an average: $$g(n)=\frac{1}{4}[g(n/4)+1]+\frac{3}{4}[g(3n/4)+1] \,.$$ Replace every term of the form $g(k)+1$ on the RHS by the average $$\frac{1}{4}[g(k/4)+2]+\frac{3}{4}[g(3n/4)+2] \,.$$ Iterate this for every $k>100$ until you express $g(n)$ as an average of exponentially many terms of the form $g(j)+\ell$, where $j \in [25,100]$ and $\ell$ is the number of iterations needed to get from $n$ to $j$. Clearly all these different values of $\ell$ will be of order $\Theta(\log n)$, so we conclude that $g(n)=\Theta(\log n)$ and $T(n)=\Theta(n\log n)$. Once you know the right order it is easy to verify.
[1] https://en.wikipedia.org/wiki/Master_theorem_(analysis_of_algorithms)