Let \(\vec f = xyz^3 + xz\)

Grad f = \(\vec \triangledown f = \left(\hat i \frac{\partial }{\partial x} + \hat j \frac{\partial}{\partial y} + \hat k \frac{\partial}{\partial z}\right)\)  \((xyz^3 + xz)\)

\(= \hat i (yz^3 + z) + \hat j(xz^3) + \hat k(3xyz^2 + x)\)

(Grad f) (1, 1, 1) = \(2\hat i + \hat j + 4\hat k\)

Given surface is \(3x^2 + y = z\)

⇒ \(3xy^2 + y - z = 0 = s\)  (Let)

Grad s = \(\vec \triangledown s = \left(\hat i \frac{\partial }{\partial x} + \hat j \frac{\partial}{\partial y} + \hat k \frac{\partial}{\partial z}\right)\)  \((3xy^2 + y - z)\)

\(= \hat i (3y^2) + \hat j(6xy +1)+ \hat k (-1)\)

(Grad s) (0, 1, 1) = \(3\hat i + \hat j - \hat k\)

Directional derivative of xyz³ + xz at (1, 1, 1) in the direction of normal to the surface 3xy² + y = z at (0, 1, 1) 

\(= (grad \,f)_{(1,1, 1)}\, . \frac{(grad f)_{(0,1,1)}}{|(grad\, f)_{(0,1,1)}|}\)

\(= (2\hat i + \hat j + 4\hat k)\,. \frac{(3\hat i + \hat j - \hat k)}{|3\hat i + \hat j - \hat k|}\)

\(= \frac{6 + 1 - 4}{\sqrt{9 + 1 +1}}\)

\(= \frac 3{\sqrt{11}}\)