Sketching the piece-wise function on paper, I found it to be of two vertical parabolas, where it opens upward in the interval x < 0, and it opens downward in the interval x >= 0.

There is no way a tangent line is to be tangent to both broken parabolas. Even at x = 0, where the y-axis is appears to be "tangent" to both broken parabolas, the y-axis, or the vertical line x=0, does not really touch the f(x) = (x+3)^2 parabola because the f(x) there is not defined.

Let us see algebraically.

There is only one tangent line, so the slope is the same everywhere on the line.

Its slope when tangent to f(x) = (x+3)^2,

m1 = f'(x) = 2(x+3)

Its slope when tangent to f(x) = -x^2 +8x -4,

m2 = f'(x) = -2x +8

m1 = m2

2(x+3) = -2x +8

2x +2x = 8 -6

x = 2/4 = 1/2

So,

m1 = 2(0.5 +3) = 7

m2 = -2(0.5) +8) = 7

The tangent line then is y = 7x +b.

Or f(x) = 7x +b -----------------------(1)

Substitute that into f(x) = (x +3)^2,

7x +b = (x +3)^2

7x +b = x^2 +6x +9

b = x^2 -x +9 ---------------------(i)

Substitute (1) into f(x) = -x^2 +8x -4,

7x +b = -x^2 +8x -4

b = -x^2 +x -4 ---------------------------(ii)

b from (i) = b from (ii),

x^2 -x +9 = -x^2 +x -4

2x^2 -2x +13 = 0

x = {2 +,-sqrt[2^2 -4(2)(13)]} / (2*2)

x = (2 +,- sqrt[-100] /4

Imaginary ??

Therefore, no such common tangent line. -------------answer.