The given linear congruence is to be solved:
$$\gcd:(3;29)=1 |17$$ $$3x+29y=1 \iff 3x \equiv 1 \pmod{29} \iff y \equiv 3^{-1} \pmod{29}$$
With the extended Euclidean algorithm one obtains: \begin{align*} 29 & = 3 \cdot 9+2\\ 3 & = 2 \cdot 1+1\\ 2 & = 2 \cdot 1 \end{align*}
By means of backward resolution: \begin{align*} 1 & = 3-2 \cdot 1\\ & = -29+10 \cdot 3\\ & = -29+10(3-29)\\ & = 10 \cdot 3-1 \cdot 29 \end{align*}
$-1 \equiv 3^{-1} \pmod{29}$
$$\implies 3x \equiv 17 \pmod{27} \iff x \equiv -1 \cdot 3x \pmod{29} \equiv -1 \cdot 17 \pmod{29}$$
Are my considerations correct so far and how do I arrive at the solution $x$?
You correctly applied the extended Euclidean algorithm to show that $1 = 10 \cdot 3 - 1 \cdot 29$. However, this does not imply that $-1$ is the multiplicative inverse of $3$ modulo $29$. Observe that $$1 \equiv 10 \cdot 3 - 1 \cdot 29 \equiv 10 \cdot 3 \pmod{29}$$ since $1$ and $10 \cdot 3$ differ by a multiple of $29$. Hence, $3^{-1} \equiv 10 \pmod{29}$.
Therefore, \begin{align*} 3x & \equiv 17 \pmod{29}\\ 10 \cdot 3x & \equiv 10 \cdot 17 \pmod{29}\\ x & \equiv 170 \pmod{29}\\ x & \equiv 5 \cdot 29 + 25 \pmod{29}\\ x & \equiv 25 \pmod{29} \end{align*}
Check: If $x \equiv 25 \pmod{29}$, then $3x \equiv 3 \cdot 25 \equiv 75 \equiv 2 \cdot 29 + 17 \equiv 17 \pmod{29}$.