$\begin{array}{l}a)\,\,y = \log_3(x^2 + 2x)\\ \text{y xác định}\,\,\Leftrightarrow x^2 + 2x > 0 \Leftrightarrow \left[\begin{array}{l}x > 0\\x < -2\end{array}\right.\\ \Rightarrow TXĐ: D = (-\infty;-2)\cap (0;+\infty)\\ b)\,\,y = \log_{0,2}(4 - x^2)\\ \text{y xác định}\,\,\Leftrightarrow 4 - x^2 > 0 \Leftrightarrow -2 < x < 2\\ \Rightarrow TXĐ: D = (-2;2)\\ c)\,\,y = \log_\sqrt2\dfrac{1}{3 - x}\\ \text{y xác định}\,\,\Leftrightarrow \dfrac{1}{3- x} > 0\Leftrightarrow 3 - x > 0\Leftrightarrow x < 3\\ \Rightarrow TXĐ: D = (-\infty;3)\\ d)\,\,y = \log_{\tfrac12}\dfrac{x+1}{x-1}\\ \text{y xác định}\,\,\Leftrightarrow \dfrac{x+1}{x-1} > 0\Leftrightarrow\left[\begin{array}{l}\begin{cases}x+1 >0 \\x - 1 >0\end{cases}\\\begin{cases}x + 1 < 0\\x -1 < 0\end{cases} \end{array}\right.\Leftrightarrow \left[\begin{array}{l}x > 1\\x < -1 \end{array}\right.\\ \Rightarrow TXĐ: D = (-\infty;-1)\cap (1;+\infty)\end{array}$