To find the second derivative of y with respect to x, we need to differentiate the function twice. Given that y = tan^2(x) + 3tan(x), let's find d^2y/dx^2 step by step.

First, let's differentiate y with respect to x to find dy/dx:
dy/dx = d/dx(tan^2(x) + 3tan(x))

To differentiate tan^2(x), we can use the chain rule. Let u = tan(x), so tan^2(x) = u^2.
Using the chain rule, du/dx = sec^2(x).
Thus, d/dx(tan^2(x)) = d/dx(u^2) = 2u * du/dx = 2tan(x) * sec^2(x) = 2tan(x)sec^2(x).

To differentiate 3tan(x), we can directly apply the derivative of tan(x), which is sec^2(x).
Thus, d/dx(3tan(x)) = 3 * sec^2(x).

Summing up both differentiations, we have:
dy/dx = 2tan(x)sec^2(x) + 3sec^2(x)

Now, let's differentiate dy/dx with respect to x to find d^2y/dx^2:
d^2y/dx^2 = d/dx(2tan(x)sec^2(x) + 3sec^2(x))

To differentiate 2tan(x)sec^2(x), we can use the product rule. Let u = 2tan(x) and v = sec^2(x).
Differentiating u with respect to x, we get du/dx = 2sec^2(x).
Differentiating v with respect to x, we get dv/dx = 2tan(x)sec(x) * tan(x) = 2tan^2(x)sec(x) = 2sec(x).

Using the product rule, we have:
d/dx(2tan(x)sec^2(x)) = u * dv/dx + v * du/dx
= (2tan(x) * 2sec(x)) + (sec^2(x) * 2sec^2(x))
= 4tan(x)sec(x) + 2sec^4(x).

To differentiate 3sec^2(x), we can directly apply the derivative of sec^2(x), which is 2sec(x)tan(x).
Thus, d/dx(3sec^2(x)) = 3 * 2sec(x)tan(x) = 6sec(x)tan(x).

Summing up both differentiations, we have:
d^2y/dx^2 = 4tan(x)sec(x) + 2sec^4(x) + 6sec(x)tan(x)

Simplifying the expression, we can factor out sec(x) from the first two terms:
d^2y/dx^2 = 2sec(x)(2tan(x) + sec^3(x)) + 6sec(x)tan(x)

Finally, factoring out sec(x) from the entire expression, we get:
d^2y/dx^2 = 2sec(x)(2tan(x) + sec^3(x) + 3tan(x))

Therefore, the correct answer is option