There's a triangle ABC like on the image below 
where the lines coming from the corners divide the opposite edge into 3 equal parts. No other information is provided (so I'd not assume that there's a right angle anywhere, despite the image suggesting that)
How to prove that the area of the piece in the middle is $9/70$ of the whole triangle? I have trouble figuring out where to start and I'd appreciate any suggestions on what to use.
Given any polygon $P_1P_2\cdots P_n$, let $[P_1P_2\cdots P_n]$ be a shorthand for its area.
First, we need a lemma. For the configuration illustrated below, we have
$$\frac{[ABD]}{[ABC]} = \frac{uv}{u+v-uv} \quad\text{ where }\quad u = \frac{AE}{AC}, v = \frac{BF}{BC}$$
To prove the lemma, let $a,b,c$ be the baricentric coordinates of $D$ with respect to $\triangle ABC$.
Notice $$\begin{cases} D \text{ lies on } AF &\implies b : c = 1-v : v\\ D \text{ lies on } BE &\implies a : c = 1-u : u \end{cases} \quad\implies\quad a : b : c = \frac1u -1 : \frac1v - 1 : 1$$ This leads to $$\frac{[ABD]}{[ABC]} = c = \frac{c}{a+b+c} = \frac{1}{\frac1u + \frac1v - 1} = \frac{uv}{u+v-uv} $$
Back to original question.
Let $\displaystyle\;\varphi(u,v) = \frac{uv}{u+v-uv}$ and label vertices as shown below.
Apply lemma to $\triangle APB, \triangle AQB, \triangle ARB, \triangle ASB$ with respect to $\triangle ABC$, we obtain
$$\frac{[APB]}{[ABC]} = \frac{[ARB]}{[ABC]} = \varphi\left(\frac13,\frac23 \right), \frac{[AQB]}{[ABC]} = \varphi\left(\frac23,\frac23\right) \quad\text{ and }\quad \frac{[ASB]}{[ABC]} = \varphi\left(\frac13,\frac13\right) $$
Notice $[PQRS] = [AQB] - [ARB] + [ASB] - [APB]$, the desired ratio is
$$\frac{[PQRS]}{[ABC]} = \varphi\left(\frac23,\frac23\right) + \varphi\left(\frac13,\frac13\right) - 2 \varphi\left(\frac13,\frac23 \right) = \frac12 + \frac15 - 2\cdot \frac27 = \frac{9}{70}$$