Hints: 1) Notice that $(1-t)x_{1}+tx_{2}-y=(1-t)(x_{1}-y)+t(x_{2}-y)$ and that norms are convex.
2) $x_{1}^{2}+3x_{2}^{4}-2x_{1}-6x_{2}=(x_{1}-1)^{2}-1+(3x_{2}^{4}-6x_{2})$ is a convex function and $\max\{f(x),0\}=\frac{f(x)+\lvert f(x)\rvert}{2}$.
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Here are some useful facts:
Any norm is convex. If $f$ is convex and $T$ is affine then $g(x) = f(T(x))$ is convex. A conic combination of convex functions is convex. A maximum of convex functions is convex.
These rules can be used to prove the functions in your question are convex.
Hints: 1) Notice that $(1-t)x_{1}+tx_{2}-y=(1-t)(x_{1}-y)+t(x_{2}-y)$ and that norms are convex.
2) $x_{1}^{2}+3x_{2}^{4}-2x_{1}-6x_{2}=(x_{1}-1)^{2}-1+(3x_{2}^{4}-6x_{2})$ is a convex function and $\max\{f(x),0\}=\frac{f(x)+\lvert f(x)\rvert}{2}$.