We have the following function
[tex]16=2xy^2+x^2y[/tex]if we derive the function with respect to x we get
[tex]\frac{d}{dx}\mleft(y\mright)=\frac{-2xy-2y^2}{x\left(x+4y\right)}[/tex]if we evaluate the result function in point 1,0 we will obtain
[tex]\frac{-2\cdot1\cdot0-2\cdot0}{1(1+4\cdot0)}=0[/tex]Thus, dy/dx at point 1,0 is 0
