Without knowing what the obviously missing symbols could be, I'm going to guess and say that the given vector field is
[tex]\mathbf F(x,y,z)=yz\,\mathbf i+xz\,\mathbf j+(xy+4z)\,\mathbf k[/tex]
though I'm not as confident about that last component as I am about the other two.
Now the goal is to find a function [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\mathbf F[/tex], which is to say [tex]f[/tex] satisfies
[tex]\begin{cases}f_x=yz\\f_y=xz\\f_z=xy-4z\end{cases}[/tex]
Here [tex]f_i[/tex] denotes the partial derivative of [tex]f[/tex] with respect to the independent variable [tex]i[/tex].
Integrating the first equation with respect to [tex]x[/tex] yields
[tex]f(x,y,z)=xyz+g(y,z)[/tex]
Differentiating this with respect to [tex]y[/tex] gives
[tex]f_y=xz=xz+g_y\implies g_y=0\implies g(y,z)=h(z)[/tex]
Differentiating with respect to [tex]z[/tex], we get
[tex]f_z=xy-4z=xy+h'(z)\implies h'(z)=-4z\implies h(z)=-2z^2[/tex]
This means you have
[tex]f(x,y,z)=xyz-2z^2[/tex]
and in particular, that [tex]\nabla f=\mathbf F[/tex], which means the line integral depends only on the value of [tex]f[/tex] at the endpoints of its path.
This means
[tex]\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=f(6,4,1)-f(3,0,-1)=24[/tex]
