We have to find f'(3) for f(x)=-5x^2, using the first principle.

Now from the first principle,

f'(3) = lim [f(x) - f(3)]/(x-3), for x->3

=> f'(3) = lim (-5x^2 + 45)/(x - 3), for x->3

=> f'(3) = lim [(-5)(x^2 - 9)]/(x-3), for x->3

=> f'(3) = lim [(-5)(x-3)(x +3)/(x-3), for x->3

=> f'(3) = (-5)*lim (x+3), for x->3

=> f'(3) = -5*(3 +3)

=> f'(3) = -5*6

=> f'(3) = -30

**The required value of f'(3) is -30**

We'll write the first principle:

lim [f(x) - f(3)]/(x-3), for x->3

lim (-5x^2 + 45)/(x - 3) = lim -5(x^2 - 9)/(x-3)

We'll recognize at numerator the difference of squares:

x^2 - 9 = (x-3)(x+3)

lim -5(x^2 - 9)/(x-3) = -5lim (x-3)(x+3)/(x-3)

We'll simplify by (x-3) and we'll get:

-5lim (x-3)(x+3)/(x-3) = -5lim(x+3)

We'll substitute x by 3 in the expression of limit and we'll get:

-5lim(x+3) = -5(3+3) = -30

**So, f'(3) = -30, using the first principle for calculating the derivative.**