Answer
$x=1+90t; y=1-90t; z=3$
Work Step by Step
The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$
Given: $f(x,y,z)=x^3+3x^2+y^3+4xy-z^3=0$
Since, we have the vector equation $r(x,y,z)=r_0+t \nabla f(r_0)$
The equation of tangent line is $v=90i -90j+0k$ or, $v=\lt 90,-90,0\gt$
Now, we have the parametric equations for $\nabla f(1,1,3)=\lt 90,-90,0\gt$ as follows:
$x=1+90t; y=1-90t; z=3+0t=3$
Thus, $x=1+90t; y=1-90t; z=3$
