Use Euclid's Algorithm to find the multiplicative inverse

Hint: $13$ and $35$ are relatively prime. Use the extended Euclidean algorithm to find the integers $x$ and $y$ such that:

$$13x + 35y = 1$$

From here, simply mod out by $35$:

$$13x \equiv 1 \pmod{35}$$

Basically you want to find $a$ such that $13a \equiv 1 \mod 35$ which is the same as: $$13a + 35k = 1, \qquad \text{For some }k\in \mathbb Z$$

Use Euclid's Algorithm on $13$ and $35$, the same way as for finding $\gcd(13,35)$.

So start with $35 = 2\cdot13 + 9$ and so on... Then substitute your answers in the line above (called reversing Euclid's Algorithm) and you'll find $a$ and $k$.

Your answer is $a$.