Exercise 3

Correct: The change in the function value is approximately equal to
$$\Delta z \approx z'_x(x_0,y_0) \Delta x + z'_y(x_0,y_0) \Delta y,$$
which means that the change in $x$ is approximately
$$\Delta x \approx \dfrac{\Delta z - z'_y(x_0,y_0) \Delta y}{z'_x(x_0,y_0)}.$$
It is given that
$$(x_0,y_0) = (1,2), \qquad \Delta y = 0.4 \qquad \text{and} \qquad \Delta z = 0.$$
The partial derivatives at $(1,2)$ are
$$\begin{align*} z'_x(x,y) &= y^2 + 3x^2y &&& z'_x(1,2) &= 2^2 + 3\cdot1^2\cdot2 = 10,\\ z'_y(x,y) &=2xy + x^3 &&& z'_y(1,2) &= 2\cdot1\cdot2 + 1^3 = 5. \end{align*}$$
Hence, the needed approximate change in $x$ is
$$\Delta x \approx \dfrac{\Delta z - z'_y(1,2) \Delta y}{z'_x(1,2)} =\dfrac{0- 5 \cdot 0.4}{10}=-0.2.$$

Go on.

Wrong: What is it you have to determine? And what is given?

See Property partial derivatives and the examples.

Wrong: What is it you have to determine? And what is given?

See Property partial derivatives and the examples.

Wrong: $\Delta x$ can certainly be determined.

See Property partial derivatives and the examples.