The line required is of the form y=ax+b where a,b are constants.

Let the original function y be represented by f(x) = x^2 - 7/(x^2)

The line required will have gradient a = f '(1). That is, the gradient is equal to the derivative of f(x) taken at the given point (1,-6).

Now for the derivative

f '(x) = 2x + 14/(x^3)

(NOTE: if you have trouble differentiating the second section, you can write it first as -7x^(-2). Multiply -7 by the power of x (-2 x -7 = 14), the power of x goes down one and becomes -3. Put it back in the denominator and there is your x^3).

Hence the gradient is given by

a = f '(1) = 2*1+14/(1^3) = 16

We now have y=16x+b.

To find b, substitute x,y for the given point (1,-6):

-6=16*1+b

=> b=-22

and we have found the tangent to f(x) at the point (1,-6) to be

y=16x-22

I hope this helps