3 Added a sentence about the initial question near the beginning.

There might be many ways to prove a variety has rational singularities.

I certainly agree with Zsolt's comment above that you should be careful with your notation, Kov\'acs theorem refers to a $Y \to X$ and above you mention $X \subset Y$, I'm assuming you are simply abusing notation.

With regards to your initial question, I would try to put $R^i \phi_* \omega_Y$ into a long exact sequence (so it depends on what $Y$ is mapping to $X$), see #3 below.

Anyway, here are some things I would try for a subvariety $X$ in $Y$ (obtained in some way)way), some of which don't use the subvariety structure.

1. Is your variety some quotient by a group action (then more info is likely needed).

2. Is $O_X$ (locally) a summand of something with rational singularities? (Apply Boutot's theorem)

3. Sandor's result, but you may as well try with a resolution $\pi : X' \to X$ and try to show that $\pi_* \omega_{X'} = \omega_X$ (at which point, this is due to Kempf, not Kovacs) (I assume you have already shown that $X$ is CM), this is easier to compute. If you have some other $Y$ with rational singularities mapping to it in some natural way, then you might be in business via Sandor's theorem. Of course, the quickest hope for computing some higher cohomology like this is sticking it in some long exact sequence.

4. If $X$ is a divisor, you could try to show that the pair $(Y, X)$ is purely log terminal (see also Lazarsfelds's book, and adjoint ideals). In this same direction, if $X$ is NOT a divisor, you could try to show that $X$ is a minimal log canonical center of some log canonical pair and then apply Kawamata's subadjunction theorem. Of course, you could also just try to show that $X$ is log terminal directly.

5. If you have specific equations, you could also try some reduction to characteristic p techniques (like things related to F-splitting and F-rationality, some of these are very effective if you have explicit equations). Even without specific equations, some of these techniques still might be useful.

6. I suppose you could also do some Bertini type tricks if somehow this subvariety is sufficiently general (for example, a general section of a base point free linear system of something with rational singularities still has rational singularities).

7. You can also see this question: http://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient/23137#23137

8. Does your variety have a small resoluation ($Y \to X$)? If it is also Cohen-Macaulay and normal, then it has rational singularities.

9. Does your variety have a Cartier divisor $D$ on it with log canonical (or maybe Du Bois) singularities such that $X \setminus D$ is has log terminal singularities (or maybe smooth). Then $X$ can be show to have rational singularities. Some things like this appeared in a paper of Koll\'ar and Shepherd-Barron (also see the related work of Karu as well as a paper of mine on Du Bois singularities).

That's all I can think of right now.

2 Tried to make #3 easier to read.

There might be many ways to prove a variety has rational singularities.

I certainly agree with Zsolt's comment above that you should be careful with your notation, Kov\'acs theorem refers to a $Y \to X$ and above you mention $X \subset Y$, I'm assuming you are simply abusing notation.

Anyway, here are some things I would try for a subvariety $X$ in $Y$ (obtained in some way)

1. Is your variety some quotient by a group action (then more info is likely needed).

2. Is $O_X$ (locally) a summand of something with rational singularities? (Apply Boutot's theorem)

3. Sandor's result, but you may as well try with a resolution $\pi : X' \to X$ and try to show that $\pi_* \omega_{X'} = \omega_X$ (at which point, this is due to Kempf, not Kovacs) (I assume you have already shown that $X$ is CM), this is easier to compute. Then this is due to Kempf I think. But if If you have some other $Y$ with rational singularities mapping to it in some natural way, then you might be in business via Sandor's theorem. Of course, the quickest hope for computing some higher cohomology like this is sticking it in some long exact sequence.

4. If $X$ is a divisor, you could try to show that the pair $(Y, X)$ is purely log terminal (see also Lazarsfelds's book, and adjoint ideals). In this same direction, if $X$ is NOT a divisor, you could try to show that $X$ is a minimal log canonical center of some log canonical pair and then apply Kawamata's subadjunction theorem. Of course, you could also just try to show that $X$ is log terminal directly.

5. If you have specific equations, you could also try some reduction to characteristic p techniques (like things related to F-splitting and F-rationality, some of these are very effective if you have explicit equations). Even without specific equations, some of these techniques still might be useful.

6. I suppose you could also do some Bertini type tricks if somehow this subvariety is sufficiently general (for example, a general section of a base point free linear system of something with rational singularities still has rational singularities).

7. You can also see this question: http://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient/23137#23137

8. Does your variety have a small resoluation ($Y \to X$)? If it is also Cohen-Macaulay and normal, then it has rational singularities.

9. Does your variety have a Cartier divisor $D$ on it with log canonical (or maybe Du Bois) singularities such that $X \setminus D$ is has log terminal singularities (or maybe smooth). Then $X$ can be show to have rational singularities. Some things like this appeared in a paper of Koll\'ar and Shepherd-Barron (also see the related work of Karu as well as a paper of mine on Du Bois singularities).

That's all I can think of right now.

1

There might be many ways to prove a variety has rational singularities.

I certainly agree with Zsolt's comment above that you should be careful with your notation, Kov\'acs theorem refers to a $Y \to X$ and above you mention $X \subset Y$, I'm assuming you are simply abusing notation.

Anyway, here are some things I would try for a subvariety $X$ in $Y$ (obtained in some way)

1. Is your variety some quotient by a group action (then more info is likely needed).

2. Is $O_X$ (locally) a summand of something with rational singularities? (Apply Boutot's theorem)

3. Sandor's result, but you may as well try with a resolution $\pi : X' \to X$ and try to show that $\pi_* \omega_{X'} = \omega_X$ (I assume you have already shown that $X$ is CM), this is easier to compute. Then this is due to Kempf I think. But if you have some other $Y$ with rational singularities mapping to it, then you might be in business via Sandor's theorem. Of course, the quickest hope for computing some higher cohomology like this is sticking it in some long exact sequence.

4. If $X$ is a divisor, you could try to show that the pair $(Y, X)$ is purely log terminal (see also Lazarsfelds's book, and adjoint ideals). In this same direction, if $X$ is NOT a divisor, you could try to show that $X$ is a minimal log canonical center of some log canonical pair and then apply Kawamata's subadjunction theorem. Of course, you could also just try to show that $X$ is log terminal directly.

5. If you have specific equations, you could also try some reduction to characteristic p techniques (like things related to F-splitting and F-rationality, some of these are very effective if you have explicit equations). Even without specific equations, some of these techniques still might be useful.

6. I suppose you could also do some Bertini type tricks if somehow this subvariety is sufficiently general (for example, a general section of a base point free linear system of something with rational singularities still has rational singularities).

7. You can also see this question: http://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient/23137#23137

8. Does your variety have a small resoluation ($Y \to X$)? If it is also Cohen-Macaulay and normal, then it has rational singularities.

9. Does your variety have a Cartier divisor $D$ on it with log canonical (or maybe Du Bois) singularities such that $X \setminus D$ is has log terminal singularities (or maybe smooth). Then $X$ can be show to have rational singularities. Some things like this appeared in a paper of Koll\'ar and Shepherd-Barron (also see the related work of Karu as well as a paper of mine on Du Bois singularities).

That's all I can think of right now.