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I start with two independent zero mean random variables $X$ and $Y$. They have variances $\sigma^2_X$ and $\sigma^2_Y$, respectively. Then I define a new random variable $$Z=0.6\,X + 0.4\, Y.$$ Is the conditional variance $V[Y|Z=z]$ constant or does $V[Y|Z=z]$ depend on $z$?

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I think in general the variance does depend on $z$. Here is an example.

$X,Y\sim\mathcal{U}[-1,1]$. Then $Z=0.6X+0.4Y\in[-1,1]$. Given $z$, $Y$ is uniform on $[-1,\frac{10}{4}z+\frac{6}{4}]$ if $z\in[-1,-\frac{2}{10}]$ and on $[\frac{10}{4}z-\frac{6}{4}]$ if $z\in[-\frac{2}{10},1]$ (just draw $[-1,1]^{2}$). If $z\in\{-1,1\}$, $Y$ has zero variance, whereas if $z\in(-1,1)$, $Y$ has strictly positive variance.

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    It is not clear to me why given $z$, $Y$ is uniform... – 2017-01-29
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    @Alik One answer is that whether $Y$ is uniform or not does not matter since the argument rests on you knowing realization of $Y$ for $z\in\{-1,1\}$, i.e., $Y$ having zero conditional variance, and you not knowing realization of $Y$ for $z\in(-1,1)$, i.e., $Y$ having strictly positive conditional variance. – 2017-01-29
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    Yes, I agree about the supports thing and the rest in your answer (just was not sure about the uniform thing). – 2017-01-29
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    @Alik another answer is this. Think about density of $(X,Y)$ on $[-1,1]^{2}$. It is uniform and equal to $\frac{1}{4}$. Now given any $z$, you know that $(X,Y)$ lie on a straight line (intersecting $[-1,1]^{2}$ with $0.6X+0.4Y=z$ and all that). And the density of those $(X,Y)$ over that line is constant. – 2017-01-29