In your picture, it looks like you're using dy/dx for the change in height, but the height y is not changing with respect to the change in the ground-distance x; it is changing with respect to the time t. So it should be "dy/dt = 8".

Also, since the balloon is rising directly vertically, then the ground-distance x is not changing; dx/dt = 0.

You have tan(@) = y/x. (I'm using "@" to stand for the angle "theta".) Differentiating with respect to time t, I believe you should get:

. . . . .sec^2(@) (d@/dt) = [(dy/dt)x - y(dx/dt)] / x^2

Since the secant is, for right triangles such as this one, equal to "(hypotenuse) over (adjacent)", use the Pythagorean Theorem to find the value of the hypotenuse when y = 25 and x = 60.

Then plug in the known values for sec(@), dy/dt, dx/dt, x, and y. Solve for d@/dt.