Show $f$ is continuous at $(0,0)$ if $f(x,y) = \frac{x^3\sin(x+y) - y^4\ln(x^2+y^2)}{x^2+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0)=0$ A function $f: \mathbb R^2  \to \mathbb R$ is defined as $$f(x,y) = \frac{x^3\sin(x+y) - y^4\ln(x^2+y^2)}{x^2+y^2}$$

if $(x,y) \neq  (0,0)$ and $f(0,0)=0$.

How can I show it is continuous at $(0,0)$? Perhaps by using the Squeeze theorem?
Note that $\displaystyle f(x,y) = \frac{x^3 \sin(x+y)}{x^2 + y^2} - \frac{y^4 \ln(x^2 + y^2)}{x^2 + y^2}$.

So it's enough to show that each of these tends to $0$.

Applying the squeeze theorem on the first

$ \displaystyle | \frac{x^3 \sin(x+y)}{x^2 + y^2}| \le \frac{|x|}{1 + \frac{y^2}{x^2}} \le |x|$.

Next $\displaystyle |\frac{y^4 \ln(x^2 + y^2)}{x^2 + y^2}| = |\frac{y^2 \ln(x^2 + y^2)}{\frac{x^2}{y^2} + 1} | \le |y^2 \ln(x^2 + y^2)|$.

Now set $x = r\cos(\theta)$ and $y = r \sin(\theta)$.  Then the limit can be estimated by $|r^2 \ln(r^2)| \to 0$ as $r \to 0$ (say by L'Hopital) independent of $\theta$.
you just need to prove $f(x,y)$ differentiable at $(0,0)$ then you can use theorem "if $f$ differentiable at $c$ then $f$ continuos at $c.$"

you can see the theorem at any analysis books