Exercise 4

Correct: The partial derivatives of $z(x,y)$ at $(x,y(x))$ and the derivative of $y(x)$ are:
$$
\begin{align*}
z'_x(x,y) &= \dfrac{1}{x^2 + y^3} \cdot 2x = \dfrac{2x}{x^2 + y^3}\\
z'_x(x,y(x)) &= \dfrac{2x}{x^2 + (10-x^2)^3}\\[2mm]
z'_y(x,y) &=\dfrac{1}{x^2 + y^3} \cdot 3y^2 = \dfrac{3y^2}{x^2 + y^3}\\
z'_y(x,y(x)) &= \dfrac{3(10-x^2)^2}{x^2 + (10-x^2)^3}\\[2mm]
y'(x) &= -2x.
\end{align*}
$$
According to the Chain rule (case 2):
$$
\begin{align*}
Z'(x) &= \dfrac{2x}{x^2 + (10-x^2)^3} + \dfrac{3(10-x^2)^2}{x^2 + (10-x^2)^3} \cdot (-2x) = \dfrac{2x}{x^2 + (10-x^2)^3} + \dfrac{-6x(10-x^2)^2}{x^2 + (10-x^2)^3} \\
&= \dfrac{2x -6x(10-x^2)^2}{x^2 + (10-x^2)^3}.
\end{align*}
$$
Finally, we plug in $x=3$:
$$
\begin{align*}
Z'(3) &= \dfrac{2\cdot 3 -6\cdot 3(10-3^2)^2}{3^2 + (10-3^2)^3} = \dfrac{-12}{10} = -\dfrac{6}{5}.
\end{align*}
$$
Go on.

Wrong: $z'_x(x,y) \neq \dfrac{1}{x^2+y^3}$ and $z'_y(x,y) \neq \dfrac{1}{x^2+y^3}$.

See Chain rule.

Wrong: Do not forget to apply the chain rule: special case 2.

See Chain rule (case 2), Example 1 and Example 2.

Wrong: It is given that $x=3$, not that $y=3$.

See Chain rule (case 2), Example 1 and Example 2.