Para determinar as derivadas parciais \(\frac{\partial z}{\partial x}\) e \(\frac{\partial z}{\partial y}\) da função \(z\) definida implicitamente pela equação \(x^3 + y^3 + z^3 + 6xyz = 1\), utilizamos a regra da diferenciação implícita. 1. Diferenciando em relação a \(x\): \[ \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial x}(y^3) + \frac{\partial}{\partial x}(z^3) + \frac{\partial}{\partial x}(6xyz) = \frac{\partial}{\partial x}(1) \] Isso resulta em: \[ 3x^2 + 0 + 3z^2 \frac{\partial z}{\partial x} + 6y \frac{\partial z}{\partial x} + 6xz = 0 \] Agrupando os termos que contêm \(\frac{\partial z}{\partial x}\): \[ 3z^2 \frac{\partial z}{\partial x} + 6y \frac{\partial z}{\partial x} = -3x^2 - 6xz \] Fatorando \(\frac{\partial z}{\partial x}\): \[ \frac{\partial z}{\partial x}(3z^2 + 6y) = -3x^2 - 6xz \] Portanto: \[ \frac{\partial z}{\partial x} = \frac{-3x^2 - 6xz}{3z^2 + 6y} \] 2. Diferenciando em relação a \(y\): \[ \frac{\partial}{\partial y}(x^3) + \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial y}(z^3) + \frac{\partial}{\partial y}(6xyz) = \frac{\partial}{\partial y}(1) \] Isso resulta em: \[ 0 + 3y^2 + 3z^2 \frac{\partial z}{\partial y} + 6x \frac{\partial z}{\partial y} + 6yz = 0 \] Agrupando os termos que contêm \(\frac{\partial z}{\partial y}\): \[ 3z^2 \frac{\partial z}{\partial y} + 6x \frac{\partial z}{\partial y} = -3y^2 - 6yz \] Fatorando \(\frac{\partial z}{\partial y}\): \[ \frac{\partial z}{\partial y}(3z^2 + 6x) = -3y^2 - 6yz \] Portanto: \[ \frac{\partial z}{\partial y} = \frac{-3y^2 - 6yz}{3z^2 + 6x} \] Resumindo: \[ \frac{\partial z}{\partial x} = \frac{-3x^2 - 6xz}{3z^2 + 6y} \] \[ \frac{\partial z}{\partial y} = \frac{-3y^2 - 6yz}{3z^2 + 6x} \]