Results of the multivariate function $f(x,y) = 3x^2 - 2xy + y^3 $ I was asked to derive 2 experessions for this multivariate function
$$
f(x,y) = 3x^2 -2xy+y^3
$$
The first is $\large{\frac{f(x+h,y) - f(x,y)}{h}}$ and the other is $\large{\frac{f(x,y+k) - f(x,y)}{k}}$
The working is as follows:
$$
\begin{align*}
\frac{f(x+h,y) - f(x,y)}{h} &= \frac{3(x+h)^2 - 2y(x+h) + y^3 -(3x^2-2xy+y^3)}{h}
\\&= \frac{3(x+h)^2 - 3x^2 -2y(x+h) + 2xy}{h}
\\&=\frac{3(x+h+x)(x+h-x) - 2y(h)}{h}
\\&=\frac{3(x+h+x)(x+h-x) - 2y(h)}{h}
\\&= 3(2x+h) -2y
\end{align*}
$$
and
$$
\begin{align*}
\frac{f(x,y+k) - f(x,y)}{k} &= \frac{3x^2 - 2x(y+k) + (y+k)^3-(3x^2-2xy+y^3)}{k}
\\&= \frac{-2x(y+k) +(y+k)^3 + 2xy-y^3}{k}
\\&=\frac{2x(-y-k+y) + (y+k)^3 - y^3}{k}
\\&=\frac{-2xk + (y+k-y)[(y+k)^2 + y(y+k) + y^2]}{k}
\\&=\frac{-2xk + k[(y+k)^2 + y(y+k) + y^2]}{k}
\\&= -2x + (y+k)^2 + y(y+k) + y^2
\end{align*}
$$
Are the calculations correct? 
UPDATE
Thanks to the community, I have verified the answer. For completeness, I shall continue to derive the partial derivatives using the above answers.
$$
\begin{align*}
\frac{\partial f}{\partial x}(x,y)=\lim_{h\to0}\frac{f(x+h,y)-f(x,y)}{h}
&= \lim_{h \to 0}3(2x+h) - 2y
\\&= 6x + 2y
\end{align*}
$$
and
$$
\begin{align*}
\frac{\partial f}{\partial y}(x,y)=\lim_{k\to0}\frac{f(x,y+k)-f(x,y)}{k}
&= \lim_{k \to 0} -2x + (y+k)^2 + y(y+k) + y^2
\\&= -2x + y^2 + y^2 + y^2
\\&= -2x + 3y^2
\end{align*}
$$

Yes. The calculations are correct.
Note that when $h\to 0$, the first expression is the partial derivative of $f(x,y)$ by $x$; and when $k\to 0$, the second expression is the partial derivative of $f(x,y)$ by $y$. 
That is, we define the partial derivative $$\frac{\partial f}{\partial x}(x,y)=\lim_{h\to0}\frac{f(x+h,y)-f(x,y)}{h}.$$
See that after getting rid of $h$ and $k$ you get the same results as a "symbolic differentiation" would have given you (i.e. $\frac d{dx}(3x^2)=6x$ and so on).
