Answer
The solutions are
a) $$\frac{f(x+\Delta x,y) - f(x,y)}{\Delta x} =6x+3\Delta x.$$
b) $$\frac{f(x,y+\Delta y) -f(x,y)}{\Delta y} =-2.$$
Work Step by Step
a) We have that
$$f(x+\Delta x,y) = 3(x+\Delta x)^2-2y = 3(x^2+2x\Delta x +\Delta x^2)-2y = 3x^2+6x\Delta x + 3\Delta x^2 - 2y.$$
This gives
$$\frac{f(x+\Delta x,y) - f(x,y)}{\Delta x}=\frac{3x^2+6x\Delta x + 3\Delta x^2 - 2y-(3x^2-2y)}{\Delta x} = \frac{3x^2+6x\Delta x + 3\Delta x^2 - 2y-3x^2+2y}{\Delta x} = \frac{6x\Delta x+3\Delta x^2}{\Delta x} = 6x+3\Delta x.$$
b) We have that
$$f(x,y+\Delta y) = 3x^2-2(y+\Delta y) = 3x^2-2y-2\Delta y.$$
This gives
$$\frac{f(x,y+\Delta y) -f(x,y)}{\Delta y} = \frac{3x^2-2y-2\Delta y-(3x^2-2y)}{\Delta y} = \frac{3x^2-2y-2\Delta y -3x^2+2y}{\Delta y} = \frac{-2\Delta y}{\Delta y} = -2.$$
