First, rearrange the equation so that every variable is on one side of the equation.
x 2 z 3 + y 2 + y sin ( π x ) = 0 {\displaystyle x^{2}z^{3}+y^{2}+y\sin(\pi x)=0}
∂ z ∂ x = − F x F z {\displaystyle {\frac {\partial z}{\partial x}}=-{\frac {F_{x}}{F_{z}}}}
We can then apply the formula above at P = ( 1 , 1 , − 1 ) {\displaystyle P=(1,1,-1)}
F x = 2 x z 3 + 0 + y π cos ( π x ) = 2 ( 1 ) ( − 1 ) 3 + ( 1 ) π cos ( π ( 1 ) ) = − 2 − π {\displaystyle F_{x}=2xz^{3}+0+y\pi \cos(\pi x)=2(1)(-1)^{3}+(1)\pi \cos(\pi (1))=-2-\pi }
F z = 3 x 2 z 2 + 0 + 0 = 3 ( 1 ) 2 ( − 1 ) 2 = 3 {\displaystyle F_{z}=3x^{2}z^{2}+0+0=3(1)^{2}(-1)^{2}=3}
As a result,
∂ z ∂ x = − F x F z = 2 + π 3 {\displaystyle {\frac {\partial z}{\partial x}}=-{\frac {F_{x}}{F_{z}}}=\color {blue}{\frac {2+\pi }{3}}}
