Помогите решить е) и ж) срочно !!!!!!!!!!

$z = {x}^{3} + {y}^{3} - 3xy \\ \\ \frac{dz}{dx} = 3 {x}^{2} + 0 - 3y = 3 {x}^{2} - 3y = 3( {x}^{2} - y) \\ \\ \frac{dz}{dy} = 3 {y}^{2} - 3x = 3( {y}^{2} - x)$
$\\ z = {e}^{ - {x}^{3} } tg(4x) \\ \\ \frac{dz}{dx} = {e}^{ { - x}^{3} } \times ( - 3 {x}^{2} ) \times tg(4x) + {e}^{ { - x}^{3} } \times ( { \frac{1}{cos(4x)} }) ^{2} \times 4 = \\ \\ = \frac{ - 3 {x}^{2} \times \cos(4x) \times \sin(4x) + 4 }{ {e}^{ {x}^{3} } \times {cos}^{2} (4x) } = \frac{ - 3 {x}^{2} \sin(8x) + 8 }{2 {e}^{ {x}^{3} } {cos}^{2} (4x)} \\ \\ \frac{dz}{dy} = 0$

Вот.