Suppose f(x,y)=4x^2+y^2, (x,y)≠0. Then,?

∇f(x,y) = <∂f/∂x, ∂f/∂y> is the vector of first partial derivatives of f.
(This extends to 3 variables and higher in a similar manner.)

Since f(x,y) = 4x^2 + y^2, we have∇f(x,y) = <8x, 2y>.
==> ∇f(3, 2) = <24, 4>.

I hope this helps!

∇f(x,y) is the gradient of the function with variables x and y which means the function f(x,y) can be located in its strongest position at ∇f(x,y). Eg: Let's say you had a graph depicting vector fields of temperatures in a room with nothing but a space heater (turned on). You would see that for whatever given multivariate function for the room temperature, the strongest position would be near the space heater.

Ok now for the math, which isn't that hard, just take the partial derivatives first with respect to x and that is your first coordinate in the vector field, then take the partial derivative with respect to y and that is your second coordinate in the vector field:

∇f(x,y)= <8x,2y>

Then for ∇f(3,2), you just plug in the coordinates.

So, ∇f(3,2)= <24,4>

Hope that helps!