[tex]f(x,y,z)=2z^2x+y^3[/tex]
[tex]f[/tex] has gradient
[tex]\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k[/tex]
which at the point (-1, 4, 3) has a value of
[tex]\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k[/tex]
I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say [tex]\vec u=15\,\vec\imath+25\,\vec\jmath[/tex], in which case we have
[tex]\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}[/tex]
Then the derivative of [tex]f[/tex] at (-1, 4, 3) in the direction of [tex]\vec u[/tex] is
[tex]D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}[/tex]
