We save the definition: $f\left(x\right)=\frac{x^2+3x-2}{x^2-5x+6}$

1) Let's perform the differentiation: $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^2+3x-2}{x^2-5x+6}\right)$

We differentiate $\mathrm{p1}$ = $x^2+3x-2$ and $\mathrm{q2}$ = $x^2-5x+6$ using: $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{p1}}{\mathrm{q2}}\right)=\frac{\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{p1}\right)·\mathrm{q2}-\mathrm{p1}·\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{q2}\right)}{{\mathrm{q2}}^2}$

2) Let's differentiate $\mathrm{p1}$: $\frac{\mathrm{d}}{\mathrm{d}x}(x^2+3x-2)$

ⓘ The derivative of a sum is the sum of the derivatives: $\frac{\mathrm{d}}{\mathrm{d}x}(x^2+3x-2)$ = $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)+\frac{\mathrm{d}}{\mathrm{d}x}\left(3x\right)+0$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)+\frac{\mathrm{d}}{\mathrm{d}x}\left(3x\right)+0$

ⓘ The derivative of x^n is n*x^(n-1): $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)$ = $2x$■Adding 0 changes nothing: $\frac{\mathrm{d}}{\mathrm{d}x}\left(3x\right)+0$ = $\frac{\mathrm{d}}{\mathrm{d}x}\left(3x\right)$■A constant factor stays in front of the derivative, (k*u)' = k*u': $\frac{\mathrm{d}}{\mathrm{d}x}\left(3x\right)$ = $3$

$2x+3$

3) Let's differentiate $\mathrm{q2}$: $\frac{\mathrm{d}}{\mathrm{d}x}(x^2-5x+6)$

ⓘ The derivative of a sum is the sum of the derivatives: $\frac{\mathrm{d}}{\mathrm{d}x}(x^2-5x+6)$ = $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)+\frac{\mathrm{d}}{\mathrm{d}x}(-5x)+0$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)+\frac{\mathrm{d}}{\mathrm{d}x}(-5x)+0$

ⓘ The derivative of x^n is n*x^(n-1): $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)$ = $2x$■Adding 0 changes nothing: $\frac{\mathrm{d}}{\mathrm{d}x}(-5x)+0$ = $\frac{\mathrm{d}}{\mathrm{d}x}(-5x)$■A constant factor stays in front of the derivative, (k*u)' = k*u': $\frac{\mathrm{d}}{\mathrm{d}x}(-5x)$ = $-5$

$2x-5$

4) Let's evaluate the resulting derivative ($\frac{\mathrm{p1}}{\mathrm{q2}}$)': $\frac{(2x+3)·(x^2-5x+6)-(x^2+3x-2)·(2x-5)}{{(x^2-5x+6)}^2}$

ⓘ Expand by multiplying each term inside the parentheses: $(2x+3)·(x^2-5x+6)$ = $2x{x}^2$$+2·x·\left(-5\right)·x$$+2x·6$$+3x^2$$+3·\left(-5\right)x$$+3·6$, $-(x^2+3x-2)·(2x-5)$ = $-(x^2·2x$$+x^2·\left(-5\right)$$+3·x·2·x$$+3x·\left(-5\right)$$-2·2·x$$-2·\left(-5\right))$

$\frac{2x{x}^2+2·x·\left(-5\right)·x+2x·6+3x^2+3·\left(-5\right)x+3·6-(x^2·2x+x^2·(-5)+3·x·2·x+3x·(-5)-2·2·x-2·(-5\left)\right)}{{(x^2-5x+6)}^2}$

ⓘ Multiplying powers with the same base adds the exponents: $x{x}^2$ = $x^3$, $x·x$ = $x^2$■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative: $2·\left(-5\right)$ = $-10$, $3·\left(-5\right)$ = $-15$■Multiplications are done before additions and subtractions: $2·6$ = $12$, $3·6$ = $18$■A minus sign before a parenthesis changes the sign of each of its terms: $-(x^2·2x$$+x^2·\left(-5\right)$$+3·x·2·x$$+3x·\left(-5\right)$$-2·2·x$$-2·\left(-5\right))$ = $-x^2·2x$$-x^2·\left(-5\right)$$-3·x·2·x$$-3x·\left(-5\right)$$+2·2·x$$+2·\left(-5\right)$

$\frac{2x^3-10x^2+12x+3x^2-15x+18-x^2·2x-x^2·\left(-5\right)-3·x·2·x-3x·\left(-5\right)+2·2·x+2·\left(-5\right)}{{(x^2-5x+6)}^2}$

ⓘ Like terms are grouped by adding their coefficients: $-10x^2+3x^2$ = $-7x^2$, $12x-15x$ = $-3x$■Multiplying powers with the same base adds the exponents: $x^{2}x$ = $x^3$, $x·x$ = $x^2$■Two minus signs cancel out: $-x^2·\left(-5\right)$ = $5x^2$■Multiplications are done before additions and subtractions: $3·2$ = $6$, $2·2$ = $4$■Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive: $-3·\left(-5\right)$ = $15$■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative: $2·\left(-5\right)$ = $-10$

$\frac{2x^3-7x^2-3x+18+2·(-x^3)+5x^2-6x^2+15x+4x-10}{{(x^2-5x+6)}^2}$

ⓘ The minus sign is placed in front of the product: $2·(-x^3)$ = $-2x^3$■The numerator and the denominator are computed before the division: $18-10$ = $8$■Like terms are grouped by adding their coefficients: $-7x^2+5x^2-6x^2$ = $-8x^2$, $-3x+15x+4x$ = $16x$

$\frac{2x^3-8x^2+16x+8-2x^3}{{(x^2-5x+6)}^2}$

ⓘ Like terms are grouped by adding their coefficients: $2x^3-2x^3$ = $0$

$\frac{0-8x^2+16x+8}{{(x^2-5x+6)}^2}$

ⓘ Adding 0 changes nothing: $0-8x^2$ = $-8x^2$

$\frac{-8x^2+16x+8}{{(x^2-5x+6)}^2}$

5) Then you can continue the derivation:

ⅰ $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^2+3x-2}{x^2-5x+6}\right)$ = $\frac{-8x^2+16x+8}{{(x^2-5x+6)}^2}$

▶Derivative:   $\frac{-8x^2+16x+8}{{(x^2-5x+6)}^2}$