How do I show that partial derivatives are continuous? The theorem says that for $f$ to be differentiable, partial derivatives of $f$ exist and are continuous.    
For example, let $f(x,y)=x^2+2xy+y^2.$ Let $(a,b)\in R^2.$
Then, I know that partial derivatives exist and $f_x(a,b)=2a+b,$ and $f_y(a,b) = a+2b$.  
In order to test the continuity, $$\lim_{(x,y)\to(a,b)}f_x(x,y)=\lim_{(x,y)\to(a,b)}2x+y=2a+b=f_x(a,b). $$
$$\lim_{(x,y)\to(a,b)}f_y(x,y)=\lim_{(x,y)\to(a,b)}x+2y=a+2b=f_y(a,b).$$
Is this the right way to check whether partial derivatives are continuous on $(a,b)$ for the multivariable case?
 A: You are on the right track but the question is, how to compute the following in general :
$$\lim_{(x,y)\to(a,b)}h(x,y).$$
We may not just do $x \to a$ then $y \to b$ (or the converse) for example. We need to consider all the manners for $(x,y)$ to tend to $(a,b)$.
Example of why it does not work :
Consider $f(x,y)= \chi_{\{x=0\} \cup \{y=0\}}$. With $\chi_A$ the function whose value is $1$ on $A$ and $0$ otherwise. Then $f$ is not continuous at $0$ but :
$$\lim_{x \to 0} f(x,0)=1=\lim_{y \to 0} f(0,y).$$
But obviously the limit for both variables does not exists, since :
$$\lim_{t \to 0} f(t,t)=0.$$
Which brings us to one of the method to show a function is not continuous at a point. Just find two different manners to tend to the point of interest such that the limit is different.
In your case the answer is clear since the partial derivative you found are polynomials. 
But one way (the most usual one I think) is to majorate the quantity $|h(a+u,b+v)-h(a,b)|$ by something who obviously tends to $0$ when $(u,v) \to 0$ or which had been proved to tend to $0$.
Finally, here is you want to prove that the function are continuous, i.e. if you consider not known that polynomials are continuous, you have to get back to definition of continuity with $\varepsilon, \delta$.
Let $\varepsilon >0$, suppose $\|(u,v)-(a,b)\|_1 \le \delta$ with $\delta = \varepsilon / 2$ then :
$$|f_x(u,v)-f_x(a,b)|=|2u+v-(2a+b)|\le 2|u-a| + |v-b| \le 2 \|(u,v)-(a,b)\|_1 \le \varepsilon$$
Then $f_x$ is continuous at $(a,b)$. I let you try for $f_y$.
