To make things easier, first rewrite
f(x) = x / (x + 7/x) = x ² / (x ² + 7)
i.e. multiply f by x/x, which is valid because x can't be 0 anyway as it's outside the domain of f.
Now, by the quotient and chain rules, we get
f '(x) = ((x ² + 7) (x ²)' - x ² (x ² + 7)') / (x ² + 7)²
… = (2x (x ² + 7) - x ² (2x)) / (x ² + 7)²
… = (2x ³ + 14x - 2x ³) / (x ² + 7)²
… = 14x / (x ² + 7)²
f ''(x) = ((x ² + 7)² (14x)' - (14x) ((x ² + 7)²)') / ((x ² + 7)²)²
… = (14 (x ² + 7)² - (14x) (2 (x ² + 7)² (x ² + 7)')) / (x ² + 7)⁴
… = (14 (x ² + 7)² - (14x) (2 (2x) (x ² + 7)²)) / (x ² + 7)⁴
… = (14 (x ² + 7)² - 4x (14x) (x ² + 7)²) / (x ² + 7)⁴
… = 14 (x ² + 7)² (1 - 4x) / (x ² + 7)⁴
… = 14 (1 - 4x) / (x ² + 7)²
… = (14 - 56x) / (x ² + 7)²