Exercise

Correct: Since the prescription of $x(y)$ cannot be determined, we use the fact that
$$x'(y) = \dfrac{1}{y'(x(y))} \qquad \text{hence} \qquad x'(1) = \dfrac{1}{y'(x(1))}.$$
In order to use this we have to determine $y'(x)$ and $x(1)$. The derivative of $y(x) =x^5 + x^3 + x + 1$ is
$$y'(x) = 5x^4 + 3x^2 + 1.$$
Moreover, note that $x(1)$ is the value of $x$ such that $y$ equals 1, hence we have to solve $y(x) = 1$:
$$\begin{align*} 1&= x^5 + x^3 + x + 1\\ 0 &= x^5 + x^3 + x = x(x^4 + x^2 + 1)\\ x = 0 &\mbox{ or } x^4 +x^2 + 1 = 0\\ &\phantom{\vee} ~~~ x^4 + x^2 = -1 \qquad \text{not possible} \end{align*}$$
Since $x^4\geq0$ and $x^2\geq0$, it also holds that $x^4+x^2\geq0$, hence $x^4 + x^2 = -1$ gives no solution. The unique solution is $x(1)=0$.
Now we can determine $x'(1)$:
$$x'(1) = \dfrac{1}{y'(x(1))} = \dfrac{1}{y'(0)} = \dfrac{1}{5\cdot0^4 + 3\cdot0^2 + 1} = \dfrac{1}{1} = 1.$$

Go on.

Wrong: The denominator of the quotient is $y'(x(1))$, not $y'(1)$.

See Derivative inverse function, Example 1 and Example 2.

Wrong: The denominator of the quotient is $y'(x(1))$, not $y'(1)$.

See Derivative inverse function, Example 1 and Example 2.

Wrong: We do not need an explicit prescription of $x(y)$ to determine $x'(1)$.

See Derivative inverse function, Example 1 and Example 2.